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COPyRIGlIT DEPOSn^ 



PRACTICAL 

MECHANICAL DRAWING 

AND MACHINE DESIGN 

SELF TAUGHT 



Drafting Tools — Geometrical Definition of Plane Figures — Properties of the Circle — Polygons — Geometrical Defini- 
tions of Solids — Geometrical Drawing — Geometrical Problems — J.Iensuration of Plane Surfaces — Mensuration of 
Volume and Surface of Solids — The Development of Curves — The Development of Surfaces — ^The Intersec- 
tion of Surfaces — Machine Drawing — Technical Definitions — Materials Used in Machine Construction 
— Shafting — Machine Design — Transmission of Motion by Belts — Horsepower Transmitted 
by Ropes — Horsepower of Gears — Transmission of Motion by Gears — Diametral 
Pitch System of Gears — Worm Gearing — Steam Boilers — Steam Engines — Tables 



BY 



CHAS. WESTINGHOUSE 



WITH OVER TWO HUNDRED ILLUSTRATIONS 




CHICAGO 

FREDERICK J. DRAKE & CO., PUBLISHERS 

1907 




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UBRARY of CONGRESS 


Two Copies Received 


J\N 9 1907 


Copyright Entry 
jDec. t>, ifoC 
CL,\SS /f^ XXC.,N». 


COPY'B, 



Copyright, 1906 

Bt Frederick J. Drake & Co. 

Chicago 



PREFACE 



If mechanical dra.wiiig is to be of any practical 
use to a person, he must be able to thoroughly un- 
derstand the form and arrangement of the various 
parts of a machine from an inspection of the 
drawings of the machine without reference to the 
machine itself. He ought also to be able to make 
drawings of a machine or the parts of a machine 
from the machine itself. As mechanical drawing 
is simply the application of the principles of ge- 
ometry to the representation of machines, a per- 
son who wishes to become thoroughly conversant 
with mechanical drawing and machine design 



should commence by studying the geometrical 
problems given in this work. The student in fol- 
lowing up the problems given, should not content 
himself by merely copying the drawings, but 
should do each example over and over, until he is 
thoroughly familiar with the principles involved 
in their construction, and also understands why 
each line is drawn. 

In working over these examples several times 
the student is not only committing them to mem- 
ory, but is at the same time becoming proficient 
in the handling of the various drawing tools. 

THE AUTHOR. 



PRACTICAL MECHANICAL DRAWING 



DRAFTING TOOLS 



Compasses. These, as well as all other instru- 
ments, should be chosen with great care on ac- 
count of their variety in shape and quality. Draft- 
ing instruments are as a rule made of German sil- 
ver and steel. The steel should be of the best 
grade and carefully tempered. The material used 
in the manufacture of some instruments is of so 




Fig. 1 — Compass 



poor a quality that it neither holds its shape nor 
wears well. It is better therefore when buying, 
to select instruments of high grade, and which are 
made of the best German silver and a good qual- 
ity of steel, as the joints are always carefully fit- 
ted and they will withstand the constant usage of 
many years, A compass set of convenient form is 



shown in Fig, 1, It has three removable parts: 
the pencil-point, the pen-point, and the lengthen- 
ing-bar. There is a hinged joint in each leg of the 
compass and the socket for the removable legs is 
provided with a clamping screw. The shanks of 
the removable legs should be a nice fit in the sock- 
et and require scarcely any effort to remove. They 







Fig. 2 — Pen and Pencil Compasses. 

should, however, stay in the socket without being 
held by the clamping screw. The lengthening-bar 
is used to extend the pen or pencil-legs when 
drawing large circles. 

A pen and pencil compass set of smaller sizei, 
without detachable legs are shown in Fig. 2, these 
instruments will be found useful in many cases 



8 



MECHANICAL DRAWING 



as a medium between the large compasses and the 
spring-bow set. 

The most important part of a pair of compasses 
is the bead, wbieb forms the hinged joint. There 
are two forms of joints: the tongue-joint, as shown 
in the left-hand view in Fig. 3, in which the head 
of one shank has a tongue, generally made of steel, 
which moves between two lugs on the other shank, 
and the pivot joint, as shown in the right-hand 
view in Fig. 3, in which each shank is reduced to 
half its thickness at the head. These shanks are 





Fig. 3 — Compass Joints. 

surrounded by a clamp or yoke, which carries two 
cone-pointed set-screws, one in each side, the 
points of these screws working in countersunks in 
the yoke. The yoke is provided with a milled or 
knurled handle to manipulate the compass. The 
head joint of the compass should move freely and 
evenly throughout its entire movement, and not 
stiff at one point and loose at another. It should 
also be tight enough in the joint to hold its adjust- 
ment when once set. Figure 4 shows the method 
of holding a compass, and the correct position of 



the fingers before and after describing a circle. 
The non-removable leg of the compass should car- 
ry a needle-point, that may be easily replaced if 
lost or damaged, and it should have a shoulder to 
prevent the point from sinking into the paper be- 
yond a certain depth. The needle-point should 
also be capable of being adjusted in or out, and 
fastened securely at any desired point, thus mak- 




Fig, 4 — Correct Manner of Holding a Compass. 

ing the leg of the compass a little longer or short- 
er as may be desired. 

The socket for the lead in the pencil-point 
should hold the lead firmly without the necessity 
of wedging it in the socket by means of paper or 
small pieces of wood. 

When first adjusting the compass for use, place 
the pen-point in the instrument and securely 
clamp it in place, firmly against the shoulder of 



DRAFTING TOOLS 



the socket, then adjust the needle-point so that its 
point is even with that of the pen. When once 
properly adjusted the needle-point should not be 
changed. The needle-point is usually made with 
a cone-point at one end and a fine shouldered- 
point at the other. The cone^point should never 
be used, as it makes too large a hole in the draw- 
ing paper. 

Hair Spring Dividers. Dividers such as are 
shown in Fig. 5 are used for laying off equal dis- 
tances and for transferring measurements from 
one part of a drawing to another, or from one 
drawing to another. They consist of steel points 




Fig. 5 — Hair Spring Dividers. 

set in German silver shanks which are hinged to- 
gether. The joints of the dividers should work 
smoothly, the legs come close together, and the 
steel points should be sharp* and of the same 
length. One of the leg^ of the dividers has a 
spring controlled by an adjustable thumb-nut. By 
means of this device, from which the instrument 
gets its name, the adjustable leg may be moved 
a trifle after the rough or approximate adjustment 
of the dividers has been made. 

Spring-bow Instruments. The spring-bow di- 
viders, pencil and pen, as shown in Fig. 6, are for 



the purpose of describing small circles and laying 
off distances of very small dimensions and are 
very convenient for these purposes. Any form of 
spring-bow instruments with interchangeable or 
removable legs will be found very unsatisfactory. 
The legs should be made of one piece of steel, to 
which the handle is attached. Any instrument in 
which the legs are separate pieces fastened to the 




6 — Spring-bow Instnunents. 



shank are undesirable, because the parts are liable 
to become loose. The spring-bow dividers are 
used like the hair-spring dividers, for the spacing 
of distances, they have the advantage of being 
fixed in any position so that there is no liability of 
a change of measurement by the handling of the 
instrument. When spacing distances the divider 



10 



MECHANICAL DRAWING 



is rotated alternately right and left, with the fore- 
finger on top of the handle. 

Ruling Pens. Ruling pens are of two different 
kinds, one kind with a hinge joint to allow the 
blades to be opened for cleaning and the other 




Fig. 7 — Ruling Pens with Hinged Blade. 

kind without a. joint and made from a solid piece 
of steel. Two sizes of ruling pens with hinged 
joints are shown in Fig. 7. The joint in this style 
of pen should be very carefully made, otherwise 
the hinged blade will very soon become loose and 
render the pen useless. The best kind of pen for 
general use is the kind shown in Fig. 8, in which 




Fig. 8 — Ruling Pen with Spring Blade. 

the upper blade springs open when the adjusting 
screw is removed from the lower blade. A pen 
such as the one just described is to be preferred to 
one with a joint, no matter how well made it may 
be. Euling pens with broad nibs and flat handles, 
as shown in Fig. 9, are preferred by many drafts- 



men, they hold a large quantity of ink and make a 
very uniform line. 




Fig. 9 — Ruling Pens with Wide Blades. 

The position in which a ruling pen should be 
held when drawing lines perpendicular to the T- 
square is shown in Fig. 10. The drawing board 




Fig. 10 — Correct Manner of Holding a Ruling Pen. 

should be placed so as to permit the light to come 
from the upper left-hand comer, this position of 
the T-square and triangle will avoid any possibil- 



DRAFTING TOOLS 



11 



ity of the shadows of the T-square blade or tri- 
angle being cast on the lines to be drawn. 

Sharpening a Ruling Pen. The blades of the 
pen should be curved at the points, and elliptical 
in shape. To sharpen the pen, screw the blades 
together and then move the pen back and forth 
upon a fine oil-stone, holding it in the position it 
should have when in use, but moving it so that the 
points are ground to the same length, and to an 




Pig. 11 — Complete Set of Instruments in Case. 

elliptical form. When this form has been secured, 
draw a folded piece of the finest emery paper two 
or three times between the blades, which are 
pressed together by the screw. This will remove 
any roughness from the inner surfaces of the 
blades, these surfaces should not be ground upon 
the oil-stone. 

When the blades are ground to the proper shape, 
they must be placed flat upon the stone and 



ground as thin as possible without giving them a 
cutting edge. To do this, the pen should be 
moved back and forth and slightly revolved at 
the same time. Both blades must be made of equal 
thickness. If either blade is ground too thin, it 
will cut the paper as would a, knife, and the pro- 
cess must be repeated from the beginning. In or- 
der to see the condition of the blades, they should 
be slightly separated while being brought to the 
proper thickness. 

Drafting Instruments. A leather-covered case 



P 



sr 



PI 



m 



Fig. 12— Slanting Blade T-Square. 

with a complete set of instruments in a velvet 
lined tray is shown in Fig. 11. This outfit is suffi- 
cient to fulfill the requirements of any ordinary 
draftsman in the way of instruments. 

T-Square. The length of a T-square is always 
measured by the length of the blade outside of the 
head. The T-square should always be as long as 
the drawing board, and if possible a little longer. 
For the general run of work the head of the T- 
square should be of a single and fixed piece, that 



12 



MECHANICAL DRAWING 



is, fastened permanently to tlie blade. The head 
should have its upper inside comer rabbeted, so 
that the guiding edge of the head may be trued up 
when occasion demands it. Al very convenient 




Fig. 13 — 60 and 45 Degree Triangles. 

form of T-square is shown in Fig. 12, which has a 
slanting blade, the working edge of which is lined 
with ebony. 




Pig. 14 — 15 Degree Triangle. 



More elaborate forms of T-squares are somei- 
times used, in which the head is double and one 
side swivels in order to draw parallel lines other 
than horizontal. The adjustable or swivel head is 



clamped in any desired position by means of a 
thumb-screw. 

Triangles or Set-Squares. Triangles are made 
of wood, hard rubber or transparent celluloid. 
The principal forms of tiiangles are shown in 



Fig. 15 — Flat Beveled-edge Scale. 

Figs. 13 and 14, which are 60°, 45° and 15° re- 
spectively. The two triangles generally used by 
draftsmen are the 60° and 45°. The former has 
angles of 30°, 60° and 90°. The latter two 45° 
and a 90° angle. 

Testing Triangles. Place the triangle on the T- 
square with the vertical edge at the right, draw a 

Fig. 16— Triangular Scale. 

fine line in contact with this edge, then reverse 
the triangle and move the vertical edge towards 
the line. If the vertical edge of the triangle and 
the line coincide the angle is 90°. If they do not 
coincide, and the vertex of the angle formed by 



DRAFTING TOOLS 



13 



the line and the vertical edge of the triangle is at 
the top, the angle is greater than 90° by half the 
angle indicated. If the, vertex of the angle is be- 
low, the angle is less than 901^by half the* amount 
indicated. 




Fig. 17 — French Curves. 



Scales. The best and most convenient form of 
scale for general use is that shown in Fig. 15. An- 
other form of scale which is very commonly used 
is shown in Fig. 16, The ordinary length of a 



scale is 12 inchesi, not counting the small portion 
at each end, which is undivided, and whose use is 
to protect the end graduations from injury. A 
scale should be used for dimensioning drawings 
only, and not used as a ruler or straight-edge. 
The measurements should be taken directly from 
the scale by laying it on the drawing, and not by 
transferring the distances from the scale to the 
drawing by means of a pair of dividers. 

Curves. For inking in lines which are neither 
straight lines nor arcs of circles, it is necessary to 




Fig. 18— Useful Form of Curve. 

use curves. They are made in a great variety of 
forms as illustrated in Fig. 17, but the form simi- 
lar to that illustrated in Fig. 18 will be found the 
most useful. They are made of wood, hard rub- 
ber and celluloid. Many curved lines can be inked 
in by means of a compass, but when the radius is 
too great, a curve should be used. 
Paper. The paper must be tough and should 



14 



MECHANICAL DRAWING 



have a surface which is not easily roughened by 
erasing lines drawn upon it. This is important 
when drawings are to be inked. For all mechan- 
ical work, the paper should be hard and strong. 

For pencil drawings a paper which is not 
smoothly calendered is best, because the pencil 
marks more readily upon an unpolished paper, 
and because its surface will not show erasures as 
quickly as that of a smooth paper. For sketching, 
several kinds of paper, which are good enough for 
the work, may be obtained both in sheets, in block 
form, and also made up in blank books. 

Whatman's paper is the best for drawings 
which are to be inked. There are two grades, hot 
and cold pressed, suitable for this use, the cold- 
pressed having the rougher surface. If the paper 
is not to be stretched, the cold-pressed is prefer- 
able, as its surface shows erasures less than that 
of the hot-pressed. The side from which the wa- 
ter-marked name is read is the right side, but 
there is little difference between the two sides of 
hot and cold pressed papers. Stretching the pa- 
per is unnecessary except when colors are to be 
applied by the brush, or when the most perfect 
inked drawing is desired. 

Pencils. Lead pencils for drafting use are 
made of different degrees of hardness and each 
kind of pencil has its grade indicated by letters 
stamped on it at one end. The grade of pencil 
mostly used by draftsmen is 4 H; a 6 H pencil is 



too hard and unless used with great care will in- 
dent the paper so that the pencil marks cannot be 
erased. A 4 H pencil requires greater care and 
more frequent sharpening, but the draftsman will 
in this manner acquire a lighter touch, which is 
of much value. Drafting pencils should always 
be sharpened tO' a chisel or wedge-shaped point, 
as shown in Fig. 19, the finishing of the point 
should always be completed with a fine file or a 
sand paper pencil sharpener, but never with a 
knife. In drawing the pencil should be held ver- 



Fig. 19 — Correct Manner of Sharpening a Pencil. 

tical, or nearly so, the arm free from the body, and 
the flat side of the chisel-point lightly touching 
the edge of the blade of the T-square. Always 
draw from left to right, or from the bottom to the 
top of the board. 

Pencil Sharpeners. Pencil sharpeners or point- 
ers are of many different kinds, from a piece of 
fine sand paper or a file to quite complicated ma- 
chines. For ordinary use a sand paper block from 
which the sheets can be removed as soon as worn 
out will be found the most convenient, as shown in 
Fig. 20. In sharpening a drafting pencil remove 



DRAFTING TOOLS 



16 



the wood from the end by means of a sharp knife, 
exposing about one-fourth to three-eighths of an 
inch of the lead. The end of the lead should then 
be sharpened to a chisel or -wedge-shaped point on 
the sand paper block. 




Fig. 20^ — Sand Paper Pencil Sharpener. 

Pencil Erasers. A pencil eraser or rubber 
should be of soft, fine-grained rubber, free from 
sand or grit and having no tendency to glaze or 
smear the surface of the drawing paper. A pencil 
eraser of the kind shown in Fig. 21 will be found 
very satisfactory for general use. 




Pig. 21— Pencil Rubber 
or Eraser. 



Fig. 22— Ink Rubber 
or Eraser. 



Ink Erasers. Inked lines should always be re- 
moved from the drawing by means of a sand-rub- 
ber, which is known as an ink eraser, but never by 
scratching the surface of the paper with a knife. 



As all drawing inks dry rapidly, and should not 
penetrate the surface of the paper, the object in 
erasing is to remove the ink from the surface of 
the paper without injury to it. An ink eraser, 
such as shown in Fig. 22, will leave the surface of 
the drawing in good condition to again receive 
ink. 

Eraser Guard. An eraser guard or shield, 
which is used to protect other lines when remov- 




Fig. 23— Erasing Guard or Shield. 

ing an mked line from the surface of the drawing 
paper, consists of a thin sheet of flexible metal, 
usually brass, provided with slots and holes of 
various shapes and sizes. The shield or guard 
permits erasures to be made of limited size with- 
out damage to the rest of the drawing. 

Drawing Ink. Black drawing ink, preferably 
some make of waterproof ink, is to be had in 



16 



MECHANICAL DRAWING 



liquid form, as shown in Fig. 24. The liquid ink 
is preferable to the Indian or Chinese stick inks, 
as shown in Fig. 25, which take considerable time 





Fig. 24— Waterproof Black l>rawing Ink. 

to prepare, besides necessitating fresh mixing 
each time the ink is used. 
Protractor. A protractor is a circular scale 




Fig. 25 — Chinese Stick Black Drawing Ink. 

and is divided into degrees and fractions of a 
degree. Protractors are made both circular and 
semi-circular in shape, the latter being the ordi- 



nary and most commonly used form, as shown 
in Fig. 26. Protractors are made of paper, horn, 
brass, German silver and steel. Protractors 
usually have their edges bevelled so as to bring 
the divisions on the scale close to the drawing 
paper. A semi-circular protractor is to be pre- 
ferred for all ordinary work. A semi-circular pro- 
tractor has a straight edge upon which the center 
of the circle is marked, so that the protractor may 




Fig. 26 — German Silver Protractor. 

be readily applied to the point at which it is 
desired to read or lay off an angle. 

Drawing Boards. A drawing board for ordi- 
nary use should be about 20 by 27 inches in size. 
The material should be of first quality clear soft 
pine, free from pitch and thoroughly kiln dried. 
The board should be made of five or six strips 
about 4 by 27 inches, well glued together and 
held from warping by two cleats on the back, as 
shown in Fig. 27. 

The working edge of the drawing board should 



DRAFTING TOOL^ 



17 



be tested from time to time, as any unevenness 
in this edge will impair the accuracy of the draw- 
ing. Some draftsmen use the lower edge of the 
board when drawing long lines parallel to the 
working edge. This necessitates making this 
•edge true, and the angle between this and the 
working edge exactly 90°. 

Thumb-Tacks. Thumb-tacks are made of Ger- 
man silver or brass disks with pointed steel pins 
in their centers. The heads or disks should have 




Fiig. 27— Drawing Board with Cleats on Back. 

very thin edges in order that the T-square may 
readily slide over them. 

Lettering Triangle. A triangle or set-square for 
laying out lettering is shown in Fig. 28. The use 
of this triangle is plainly indicated by its name. 

Section Liners. A section liner is a device for 
drawing a series of parallel lines equi-distant 
from each other. One form of section liner is 
shown in Fig. 29. Its operation is as follows: 
Place the instrument in the position shown in the 



drawing, and rule a line along its vertical edge. 
Hold the straight-edge firmly in place, and slide 
the triangle along it until the other side of the 
tapered edge of the tongue comes in contact with 
the other stud and holds it in this position, then 
allow the straight-edge to be drawn forward by 
the spring. Then draw a second line which will 




Fig. 28— Lettering Triangle. 

evidently be parallel to the first. The distance 
between the lines is regulated by moving the 
tongue in or out between the studs, as far as de^ 
sired. 

Another form of section liner is shown in Fig. 
30, having a horizontal instead of a vertical ad- 
justment to regulate the width of the spacing. 

Beam Compasses. A beam compass is jiot, as a 



18 



MECHANICAL DRAWING 



rule, included in a draftsman's outfit, but every 
well equipped drafting room should have one. A 
beam compass is shown in Fig. 31, with removable 




Pig. 29 — Section Liner with Vertical Adjustment. 

legs and pen, pencil and needle-points. The right- 
hand leg in the illustration has a horizontal ad- 




Fig. 30'— Section Liner with Horizontal Adjustment. 

justment of about one-half an inch, operated by 
the milled thumb-nut shown. 

Water Colors. These may be obtained in the 
form of a thick paste in small porcelain pans, or 



in thin paste or semi-liquid form in collapsible 
tubes. The colors in tubes are liable to get hard, 
in which case they cannot be expelled from the 
tubes by pressure. The caps to the tubes also get 
stuck in place by the colors and are often removed 




Fig. 31 — Beam Compass with Pen, Pencil and Needle-Points. 



with much difficulty. For the draftsman's pur- 
poses the moist colors in pans will be found the 
most satisfactory. 

A box of moist water colors is illustrated in 
Fig. 32. The colors should be kept in a box of 
this kind, which can also be used as a palette. 



DRAFTING TOOLS 



19 



The box keeps all dust and dirt from the colors 
and prevents them from drying out rapidly. 
Water Color Brushes. These are made from 




Pig. 32— Moist Water Colors in Case. 

black or red sable and camel's hair. Black sable 
brushes are too expensive for the draftsman's 
ordinary use. 
The best grade of camel's hair brushes such as 



are shown in Fig. 33, will be found quite satis- 
factory for ordinary use. 

To ascertain whether a brush is of good qual- 
ity or not, dip it in water until thoroughly wet, 



rig. 33— Water Color Brushes. 

and then remove the water from it by an quick 
motion. The brush if of good quality should as- 
sume a convex shape, come to a fine point and also 
preserve its elasticity. 



GEOMETRICAL DEFINITIONS OF PLANE FIGURES. 



A line is the "boundaiy or limit of a surface, 

A line has only one dimension, that of length. 

A point is considered as the extremity or limit 
of a line. The place where two lines intersect is 
also a point. A point has position hut no dimen- 
'siona. 

Jn practical work a point is represented by a 
fine dot. 

Lines may be either straight, broken or curved. 

A straight line is one which has the same direc- 
tion throughout its length. 

A straight line is also called a right line. 

A straight line is usually called a line simply, 
and when the word line occurs it is to be under- 
stood as meaning straight line unless otherwise 
specified. A straight line is the shortest distance 
between two points. If any other path between 
the points were chosen, the line would become 
curved or broken. Therefore two points deter- 
mine the position of the straight line joining 
them. 

A broken line is one which changes direction 
at one or more points. 

A curved line is one which changes direction 



20 



constantly throughout its length. The word curve 
is used to denote a curved line. 

Lines may be represented as full, dotted, 
dashed, or dot-and-dashed. 

A full line is one which is continuous through- 
out its length. 

A dotted line is one which is composed of alter- 
nate dots and spaces. 

A dash line is one which is composed of alter- 
nate dashes and spaces. 

A dot-and-dash line is one which is composed 
of dots, spaces and dashes. These may be ar- 
ranged in several ways according to the character 
of the line, that is, the meaning it is to convey. 

Surfaces may be either plane or curved. A 
plane surface is usually called a plane. 

A plane is such a surface that if a straight line 
be applied to it in any direction, the line and the 
surface will touch each other throughout their 
length. 

A curved surface is one no part of which is a 
plane. 

Any combination of points, lines, surfaces or 
solids is termed a figure. 



GEOMETRICAL DEFINITIONS 



21 



A plane figure is one wMch lias all of its points 
in the same plane. 

Plane geometry treats of figures whose points 
all lie in the same plane. 

Lines may be so situated as to be parallel or 
inclined to each other. 

Parallel lines are those which have the same or 
opposite directions. Parallel lines are every- 
where equally distant. Parallel lines will not 
meet, however far produced. 

Inclined lines are those other than parallel. In- 



rig. 34. 



Fig. 35. 



clined lines will always meet if produced far 
enough. Their mutual inclination forms an angle. 

The extremities of a surface are liaes. 

A plane rectilineal angle is the inclination of 
two straight lines to one another in a plane which 
meet together, but are not in' the same straight 
line as in Fig. 34. 

When -a straight line, standing on another 
straight line, makes the adjacent angles equal to 
one another, each of the angles is called a right 



angle and the straight line which stands on the 
other is called a perpendicular to it as in Fig. 35. 
An obtuse angle is that which is greater than a 
right angle as in Fig. 36. 



rig. 36. 

An acute angle is that which is less than a 
right angle as in Fig. 34. 

A term or boundary is the extremity of any- 
thing. 

An equilateral triangle is that which has three 
equal sides as in Fig. 37. 





Fig. 37. 

An isosceles triangle is that which has two sides 
equal as in Fig. 38. 

A scalene triangle is that which has three un- 
equal sides as in Fig. 39. 



22 



MECHANICAL DRAWING 



A right angled triangle is that which has a 
right angle as in Fig. 40. 

An obtuse-angled triangle is that which has an 
obtuse angle as in Fig. 39. 




The hypothenuse in a right angled triangle is 
the side opposite the right angle as in Fig. 40. 

A square is that which has all its sides equal 
and all its angles right-angled as in Fig. 41. 

A rectangle is that which has all its angles 



Fig. 41. 



Fig. 42. 



right angles, but only its opposite sides equal as 
in Fig. 42. 

A rhombus is that which has all its sides equal, 
but its angles are not right angles as in Fig. 43. 

A quadrilateral figure which has its opposite 



sides parallel is called a parallelogram as in Figs. 
41, 42 and 43. 

A line joining two opposite angles of a quadri- 
lateral is called a, diagonal. 

An ellipse is a plane figure bounded by one con- 
tinuous curve described about two points, so that 
the sum of the distances from every point in the 
curve to the two foci may be always the same — 
Fig. 44. 




Fig, 43. 



Fig. 44. 



PROPERTIES OF THE CIRCLE. 

A circle contains a greater area than any other 
plane figure bounded by the same length of cir- 
cumference or outline. 

A circle is a plane figure contained by one line 
and is such that all straight lines drawn from a 
point within the figure to the circumference are 
equal, and this point is called the center of the 
circle. 

A diameter of a circle is a straight line drawn 
through the center and terminated both ways by 
the circumference, as AC in Fig. 45. 



GEOMETEICAL DEFINITIONS 



23 



A radius is a straight line drawn from the cen- 
ter to the circumference, as LH in Fig. 45. 

A semicircle is the figure contained by a diam- 
eter and that part of circumference cut off by a 
diameter as AUG in Fig. 45. 

A segment of a circle is the figure contained by 
a straight line and the circumference which it 
cuts off, as DHE in Fig. 45. 

A sector of a circle is the figure contained by 
two straight lines drawn from the center and the 




Pig. 45. 

circmnference between them, as LMO in Fig. 45. 

A chord is a straight line, shorter than the 
diameter, lying within the circle, and terminated 
at both ends by the circumference as DE in 
Fig. 45. 

An arc of a circle is any part of the circumfer- 
ence as DHE in Fig. 45. 

The versed sine is a perpendicular joining the 



middle of the chord and circumference, as GH in 
Fig. 45. 

Circumference. Multiply the diameter by 3.1416, 
the product is the circumference. 

Diameter. Multiply the circumference by 
.31831, the product is the diameter, or multiply 
the square root of the area by 1.12837, the prod- 
uce is the diameter. 

Area. Multiply the square of the diameter by 
.7854, the product is the area. 




fig. 46. 

Side of the square. Multiply the diameter by 
.8862, the product is the side of a square of equal 
area. 

Diameter of circle. Multiply the side of a 
square by 1.128, the product is the diameter of a 
circle of equal area. 

To find the versed sine, chord of an arc or the 
radius when any two of the three factors are 
given— Fig. 46. 



24 



MECHANICAL DRAWING 



E = 



C2+4V2 



8V 



C=2i/V(2E— V) 






'4R2_C2 



To find the length of any line perpendicular to 
the chord of an arc, when the distance of the line 
from the center of the chord, the radius of the arc 
and the length of the versed sine are given — 
Fig. 47. 

I* c >i 




rig 47. 



N=v'(R2_X2)— (K— H 



R= 



C2+4V2 



8V 



C=2v/V(2R— V) 



i 
V=R-y 



I4R2— 02 



To find the diameter of a circle when the chord 
and versed sine of the arc are given. 

DG2+GH2 



AC=- 



To find the length of any arc of a circle, when 
the chord of the whole arc and the chord of half 
the arc are given— Fig. 48. 




8DH— DE 



GH 



Arc DHE= 



Rg. 48. 

jA Tangent is a straight line which touches the 
circumference but does not intersect it. The point 
where the tangent touches the circle is called the 
Point of Tangency. 

Two Circumferences are tangent to each other 
when they are tangent to a straight line at the 
same point. 

A Secant is a straight line which intersects the 
circumference in two points. 

A Polygon is inscribed in a circle when all of 
its sides are chords of the circle. 

A Polygon is circumscribed about a circle 
when all of its sides are tangent to the circle, and 
a circle is circumscribed about a polygon when the 
circumference passes through all the vertices of 
the polygon. 



GEOMETRICAL DEFINITIONS 



25 



DEFINITION OF POLYGONS. 

A polygon, if its sides are equal, is called a 
regular polygon, if unequal, an irregular polygon. 
A pentagon is a five-sided figure. 
A hexagon is a six-sided figure — Fig. 49. 
A heptagon is a seven-sided figure. ' 



Fig. 49 — Hexagon. 

An octagon is an eight-sided figure. 
A nonagon is a nine-sided figure. 
A decagon is a ten-sided figure. 
A unadecagon is an eleven-sided figure. 
A duodecagon is a twelve-sided figure. 



GEOMETRICAL DEFINITION OF SOLIDS. 

A solid has length, breadth and thickness. The 
boundaries of a solid are surfaces. 

A solid angle is that which is made by two or 
more plane angles, which are not in the same 
plane, meeting at one point. 

A cube is a solid figure contained by sis equal 
squares — ^Fig. 50. 

A prism is a solid figure contained by plane 



figures of which two that are opposite are equal, 
similar, and parallel to one another, the other 
sides are parallelograms — ^Fig. 51. 

A pyramid is a solid figure contained by planes, 
one of which is the base, and the remainder are 




Fig. 50— Cube. 



Fig. 51 — Frism. 



triangles, whose vertices meet a point about the 
base, called the vertex or apex of the pyramid — 
Fig. 52. , - 





Fig. 52 — Psrramid. 



Fig. 53— Cylinder. 



A cylinder is a solid figure described by the 
revolution of a rectangular or parallelogram about 
one of its sides — ^Fig. 53, 



26 



MECHANICAL DRAWING 



The axis of a cylinder is the fixed straight line 
about which the parallelogram revolves. 

The ends of a cylinder are the circles described 
by the two revolving sides of the parallelogram. 





Sphere. 



A sphere is a solid figure described by the rev- 
olution of a semicircle about its diameter, which 
remains fixed — Fig. 54, 

The axis of a sphere is the fixed straight line 
about which the semicircle revolves. 



The center of a sphere is the same as that of 
the semicircle. 

The diameter of a sphere is any straight line 
which passes through the center and is terminated 
both ways by the surface of the sphere. 

A cone is a solid figure described by the revolu- 
tion of a right-angled triangle about one of its: 
sides containing the right angle, which side re- 
mains fixed — Fig, 55. 

The axis of a cone is the circle described by that 
side of the triangle containing the right angle 
which revolves. 

The base of the cone is the circle described by 
that side of the triangle containing the right angle 
which revolves. 

If a cone be cut obliquely so as to preserve the 
base entirely, the section is an ellipse. 

When a cone is cut by a plane parallel to one 
of sloping sides, the section is a parabola, if cut 
at right angles to its base, an hyperbola. 



MECHANICAL DRAWING 



While many draftsmen are familiar with all of 
the problems given in this section of the work, 
it is not to be expected that all draftsmen or stu- 
dents are thoroughly conversant with all of them, 
and it is intended that this section of the work 
shall be used not only as reference data but for 
practical examples of elementary mechanical 
drawing. If the different problems given in this 
section are drawn with great accuracy, the tech- 
nical skill acquired in drawing and proper han- 
dling of the different instruments will be found to 



be of great value. It will not be necessary to ink 
in these simple geometrical problems, as it is bet- 
ter to acquire precision or accuracy in pencil 
work before going further. These problems are 
believed to be an essential part of a work on me- 
chanical drawing. To' understand geometry cer- 
tain qualities of mind are absolutely necessary, 
and many persons find it impossible to grasp even 
the simple problems of this study. The draftsman 
or student who is without practical knowledge of 
geometry is very poorly equipped for his duties. 



27 



GEOMETRICAL PROBLEMS 



THE CONSTRUCTION OF ANGLES. 

To bisect a given angle. Let DAG be the given 
angle. With center A and any radius AE de- 
scribe an arc cutting AC and AD at E and U. 
With the same radius and centers E and G, de- 
scribe arcs intersecting at H, and join AH. The 
angle DAC is bisected — ^Fig. 56. 

To construct an angle of 30°. With radius AE 





rig. 57. 



and with center A and E, describe arcs inter- 
secting at G. With the same radius and with 
centers E and G, describe arcs intersecting at D, 
and join AD. The angle DAC contains 30°— 
Fig. 57. 

To construct an angle of 60°. With radius AE, 
and with centers A and E, describe arcs intersect- 



ing at G, draw AD through G. The angle DAG 
contains 60°— Fig. 58. 

To construct an angle of 45°. With radius AE 
and centers A and E, describe arcs intersecting at 
F, draw EG through F, and make FG equal to 
FE. Join GK, and with center ^ and radius AE 
make AH equal to AE, with the same radius and 
with centers E and H describe arcs intersecting at 





L, draw AD through L. The angle DAC is 45° — 
Fig. 59. 

To construct an angle of 90°. With radius AE 
and centers A and E, describe arcs intersecting at 
F, with the same radius and center F describe the 
arc AGD, with radius AE, lay off AG and GD 
and join DA. The angle DAG is 90°— Fig. 60. 



28 



GEOMETEICAL PROBLEMS 



29 



To bisect a straight line — Fig. 61. Let BC be 

the straight line to be bisected. With any con- 
venient radius greater than A3 or AC describe 




xF 



\ 



H- 



Fig. 60. 

arcs cutting each other at D and E. A line drawn 
through D and E will bisect or divide the line BC 
into two equal parts. 




Fig. 61. 



Fig. 62. 



To erect a perpendicular line at or near the end 
of a straight line — Fig. 62. With any convenient 
radius and at any distance from the line AC, de- 



scribe an arc of a circle as ACE, cutting the line 
at A and C. Through the center R of the circle 
draw the line ARE, cutting the arc at point E. A 
line drawn from C to E will be the required per- 
pendicular. 

To divide a straight line into any number of 
equal parts — Fig. 63. Let AB be the straight line- 
to be divided into a certain number of equal parts : 
From the points A and B, draw two parallel lines 
AD and BC, at any convenient angle with the line 
AB. Upon AD and BC set off one less than the 




A D ' G' c 
Fig. 64. 



number of equal parts required, as A-1, 1-2, 2-D, 
etc. Join C-1, 2-2, 1-D, the line AB will then be 
divided into, the required number of equal parts. 

To find the length of an arc of a circle^ — Fig. 64. 
Divide the chord AC of the arc into four equal 
parts as shown. With the. radius AD equal to 
one-fourth of the chord of the arc and with A as 
the center describe the arc DE. Draw the line EG 
and twice its length will be the length of the arc 
AEC. 

To draw radial lines from the circumference of 



30 



MECHANICAL DRAWING 



a circle when the center is inaccessible — Fig. 65. 

Divide the circumference into any desired number 
of parts as AB, BC, CD, DE. Then with a radius 




Fig. 65. 




greater than the length of one part, describe arcs 
cutting each other as A-2, C-2, B-3, D-3, etc., also 
B-1, D-5. Describe the end arcs A-1, E-5 with a 



radius equal to B-2. Lines joining A-1, B-2, C-3, 
D-4 and E-5 will all be radial. 
To inscribe any regular polygon in a circle — 

Fig. 66. Divide the diameter AB of the circle into 
as many equal parts as the polygon is to have 
sides. With the points A and B as centers and 
radius AB, describe arcs cutting each other at C. 
Draw the line CE through the second point of di- 
vision of the diameter of AB, intersecting the cir- 




cumference of the circle D. A line drawn from 
B to D is one of the sides of the polygon. 

To cut a beam of the strongest shape from a cir- 
cular section — Fig. 67. Divide any diameter CB 
of the circle into three equal parts as CF, FE and 
EB. At E and F erect perpendiculars EA and FD 
on opposite sides of the diameter CB. Join AB, 
BD, DC and CF. The erect angle ABCD will be 
the required shape of the beam. 



GEOMETRICAL PROBLEMS 



31 



To divide any triangle into two parts of equal 
area — Fig. .68. Let ABC be the given triangle: 
Bisect one of its sides AB at D and describe the 
semicircle AEB. At D erect the perpendicular DE 
and with center B and radius BE describe the arc 
EF which intersects the line AB at F. At F draw 
the line AG parallel at AC, this divides the tri- 
angle into two parts of equal area. 

To inscribe a circle of the greatest possible di- 




Fig. 68. 



Fig. 69. 



ameter in a given triangle — Fig. 69. Bisect the 
angles A and B, and draw the lines, AD, BD which 
intersect each other at D. From D draw the line 
CD perpendicular to AB. Then CB will be the 
radius of the required circle CEF. 



To construct a square equal in area to a given 
circle — Fig. 70. Let ACBD be the given circle: 
Draw the diameters AB and CD at right angles 
to each other, then bisect the half diameter or 
radius DB at E and draw the line FL, parallel 
to BA. At the points C and F erect the perpen- 
diculars CH and FG, equal in length to CF. Join 
HG, then CFGH is the required square. The 





Fig. 70. 



dotted line FL is equal to one-fourth the circle 
ACBD. 

To construct a rectangle of the greatest possible 
area in a given triangle — Fig. 71. Let ABC be the 

given triangle: Bisect the sides AB and BC at G 
and F. Draw the line GD and from the points G 
and D, draw the lines GF and DE perpendicular 
to GD, then EFGD is the required rectangle. 



32 



MECHANICAL DEAWING 



To construct a rectangle equal in area to a given ; 
triangle— Fig. 72. Let ABC be the given triangle: 
Bisect the base AB of the triangle at D and erect 
the perpendiculars DE • and BF at D and B. 
Through C draw the line EOF intersecting the 




perpendiculars DE and B at E and F. Then 
BDEF is the required rectangle. 

To construct a triangle equal in area to a given 
parallelogram — Fig. 73. Let ABCD be the given 
parallelogram: Produce the line AB at B and 

D C 




make BE^ equal to AB. Join the points A and 
and ACE will be the triangle required. 

To inscribe a square within a given circles- 
Fig. 74. Let ADBC be the given circle: Draw 



the diameters AB and CD at right angles to each 
other. Join AD, DB and CA, then ACBD is the 
inscribed square. 

To describe a square without a given circle — 
Fig. 75. Draw the diameters AB and CD at right 
angles to each other. Through A and B draw the 
lines EF and GH, parallel to CD, also draw the 
lines EGr and FH through the points C and D and 




E^ A 

C 



Fig. 74. 



B 
Fig. 75. 



H 



parallel to' AB, this completes the required square 
EFGH. 

To construct an octagon in a given square — 
Fig. 76. Let ABCD be the given square: Draw 
the diagonal lines AC and BD, which intersect 
each other at the point 0. With a radius equal to 
AG or OC, describe the arcs EF, GH, IK and LM. 
Connect the points EK, LG, FI and HM, then 
GFIHMKEL is the required octagon. 



GEOMETRICAL PROBLEMS 



33 



To construct a circle equal in area to two given 
circles — Fig. 77. Let AB and AC equal the diam- 
eters of the given circles : Erect AC at A and at 
right angles to AB. Connect B and C, then bisect 
the line BC at D and describe the circle ACB 
which is the circle required and is equal in area 
to the two given circles. 

To describe an octagon about a given circle — 
Fig. 78. Let ACBD be the given circle: Draw 




Fig. 77. 



the diameters AB and CD at right angles to each! 
other. With any convenient radius and centers 
A, C, B and D describe arcs intersecting each 
other at E, H, F and G-. Join EF and GH which 
form two additional diameters. At the points AB 
and CD draw the lines KL, PR, MN and ST par- 
allel with the diameters CD and AB respectively. 
At the points of intersection of the circumference 
of the circle by the lines EF and GH, draw the 



lines KP, RM, NT and SL parallel with the lines 
EF and HG respectively, then PRMNT'SLK is the 
required octagon. 

To draw a straight line equal in length to a 
given portion of the circumference of a circle — 
Fig. 79. Let ACBD be the given circle: Draw the 
diameters AB and CD at right angles with each 
other. Divide the radius RB into four equal parts. 
Produce the diameter AB and B and make BE 




equal to three of the four parts of RB. At A draw 
the line AF parallel to CD and then draw the line 
ECF which is equal to one-fourth of the circum- 
ference of the circle ACBD. If lines be drawn 
from E through points in the circumference of the 
circle as 1 and 2, meeting the line AF and G and H, 
then C-1, 1-2 and 2-A will equal EG, GH and HA 
respectively. 
To construct a square equal in area to two given 



34 



MECHANICAL DRAWING 



squares — Fig. 80. Let AC and AD be the length 
of the sides of the given squares : Make AD per- 
pendicular to AC and connect DC, then DC is one 
of the sides of the square DCEG which is equal to 
the two given squares. 

To inscribe a hexagon in a given circle — Fig. 81. 
Draw a diameter of the circle as AB: With cen- 
ters A and B and radius AC or BGr, describe arcs 




cutting the circumference of the circle at D, B, 
F and G. Join EF, FB, BG, GD, DA and AE, this 
gives the required hexagon. 

To describe a cycloid, the diameter of the gen- 
erating circle being given — Fig. 82. Let BD be 
the generating circle: Draw the line ABC equal 
in length to the circumference of the generating 



circle. Divide the circumference of the generating 
circle into 12 parts as shown. Draw lines from 
the points of division 1, 2, 3, etc., of the circum- 




Fig. 81. 



ference of the generating circle parallel to the 
line ABC and on both sides of the circle. Lay off 
one division of the generating circle on the lines 




5 and 7, two divisions on the lines 4 and 8, three 
divisions on the lines 3 and 9, four divisions on 
the lines 2 and 10, and five divisions on the lines 



GEOMETRICAL PROBLEMS 



35 



1 and 11. A line traced through the points thus 
obtained will be the cycloid curve required. 

To develop a spiral with uniform spacing — Fig. 
83. Divide the line BE into as many equal parts 
as there are required turns in the spiral. Then 
subdivide one of these spaces into four equal 
parts. Produce the line BE to 4, making the ex- 
tension E-4 equal to two of the subdivisions. At 



1 draw the line 1-D, lay off 1-2 equal to one of 
the subdivisions. At 2 draw 2-A perpendicular 
to 1-D and at 3 in 2-A draw 3-C, etc. With center 

1 and radius 1-B describe the arc BD, with center 

2 and radius 2-D describe the arc DA, with center 

3 and radius 3-A, etc., until the spiral is com- 
pleted. If carefully laid out the spiral should ter- 
minate at E as shown in the drawing. 




MENSURATION 



Mensuration is that branch of arithmetic which 
is used in ascertaining the extension and solidity 
or capacity of bodies capable of being measured. 

DEFINITIONS OF ARITHMETICAL SIGNS. 



6- 



the Sum. 
-3=3, the 



Re- 



as 8X4=32, the 



=4^*=4. 



=Sign of Equality, as 4+8=12. 

+Sign of Addition, as 6+6=12, 

— Sign of Subtraction, as 
mainder. 

X Sign of Multiplication, 
Product. 

^-Sign of Division, as 24-^-6= 

V Sign of Square Eoot, signifies Evolution or 
Extraction of Square Root. 

2 Sign of to be Squared, thus 82=8X8=64. 

3 Sign of to be Cubed, thus 33=3x3x3=27. 

MENSURATION OF PLANE SURFACES. 

To find the area of a circle — Fig. 84. Multiply 
the square of the diameter by .7854. 

To find the circumference of a circle. Multiply 
the diameter by 3.1416. 
Circle: Area^ .78540^ 
Circ.=3.1416D 



To find the area of a semi-circle —Fig 85. 

tiply the square of the diameter by .3927. 

To find the circumference of a semi-circle. 
tiply the diameter by 2.5708. 
Semi-Circle: Area^ .39270^ 
Circ.=2.5708D 



Mul- 



Mul- 




D >< 

Fig. 84— Circle. 




P 3 

85 — Semi-circle. 



To find the area of an annular ring— Fig. 86. 

From the area of the outer circle subtract the 
area of the inner circle, the result will b© the area 
of the annular ring. 

To find the outer circumference of an annular 
ring. Multiply the outer diameter by 3.1416. 

To find the inner circumference of an annular 
ring. Multiply the inner diameter by 3.1416. 



36 



MENSURATION 



37 



Annular ring; Areap= .7854 (D^— H^) 
Out. circ.=3.1416 D 
Inn. circ.=3.1416 H 
To find the area of a flat-oval — Fig. 87. Mul- 

1 

I 
I 
I 

.r 
» 
I 
I 




I- 

Fig. 



-Annular Ring. 



Fig, 



U— D- 
87— Flat Oval. 



tiply the length by the width and subtract .214 
times the square of the width from the result. 
To find the circumference of a flat-oval. The 




Fig. 88— Paraljola. 



I I 

k D *l 

Fig. 89— Square. 



circumference of a flat-oval is equal to twice its 
length plus 1.142 times its width. 
Flat-oval: Area^D (H— 0.214D) 
Circ.=2 (HX0.571D) 



To find the area of a parabola— Fig. 88. Mul- 
tiply the base by the height and by .667. 
Parabola: Area;=.667(DxH) 

To find the area of a square — Fig. 89. Multiply 
the length by the width, or, in other words, the 
area is equal to square of the diameter. 

To find the circumference of a square. The cir- 
cumference of a square is equal to the sum of the 
lengths of the sides. 
Square: Area^D^ 
Circ.=4D 




1* M — H 

Fig. 90 — Rectangle. 




l« D 

Fig. 91 — Parallelogram. 



To find the area of a rectangle — Fig. 90. Mul- 
tiply the length by the width, the result is the area 
of the rectangle. 

To find the circumference of a rectangle. The 
circumference of a rectangle is equal to twice the 
sum of the length and width. 
Rectangle: Area^=DxH 

Circ.=2(DxH) 

To find the area of a parallelogram — Fig. 91. 
Multiply the base by the perpendicular height. 
Parallelogram : Area^=D x H 



38 



MECHANICAL DRAWING 



To find the area of a trapezoid— Fig. 92. Mul- 
tiply half the sum of the two parallel sides by the 
perpendicular distance between the sides. 

„ ., . (HE+D) 

Trapezoid: Area= -^^ — ^ — - 

To find the area of an equilateral triangle — Fig. 
93. The area of an equilateral triangle is equal to 
the square of one side multiplied by .433. 

To find the circumference of an equilateral tri- 
angle. The circumference of an equilateral tri- 





k D >i 

Fig. 92— Trapezoid. 



Fig. 



U D *i 

93 — Equilateral Triangle. 



angle is equal to the sum of the length of the 
sides. 

Equilateral triangle: Area^.433D2 

Circ.=3D 

To find the area of a right-angle or an isosceles 
triangle — Fig. 94. Multiply the base by half the 
perpendicular height. 

To find the circumference of any regular poly- 
gon — ^Fig. 95. The circumference of any polygon 
is equal to the sum of the length of the sides. 



Polygon: Area^^ 



No. of sides XDXP 



Circ.=No. of sides XD 
D=Length of one side. 
P=Perpendicular distance from the 
center to one side. 




Ue — D — >J u D *^ 

Fig. 94 — Right-angle and Isosceles Triangles. 

To find the area of an ellipse — Fig. 96. Mul- 
tiply the long diameter by the short diameter and 
by .7854. 





Fig. 96— Ellipse. 



Uf D — -J 

Fig. 95 — Regular Polygon. 

To find the circumference of an ellipse. Mul- 
tiply half the sum of the long and short diameters 
by 3.1416. 



MENSURATION 



39 



Ellipse: Area^ .7854 (DxH) 
Circ.=1.5708 (D+H) 

To find the area of a hexagon — Fig. 97. Mul- 
tiply the square of one side by 2.598. 
To find the circumference of a hexagon. The 

circumference of a hexagon is equal to the length 
of the sides. 



Hexagon: 



Area=2.598 S^ 
Circ.=6 S 
r)=1.732 S 



1 
I 

(0 

I 

_i 



to 
I 

1. 



U- D j4 

Fig. 97— Hexagon. 



U- D— J 

Fig. 98— Octagon. 



To find the area of an octagon — Fig. 98. Mul- 
tiply the square of the short diameter by .828. 

To find the circumference of an octagon. The 
circumference of an octagon is equal to the sum 
of the length of the sides. 

Octagon: Area^.828 D^ 
Circ.=8 S 
^=.414 P 



MENSURATION OF VOLUME AND SURFACE 
OF SOLIDS. 

To find the cubic contents of a sphere — Fig. 99. 

Multiply the cubic of the diameter by .5236. 

To find the superficial area of a sphere. Mul- 
tiply the square of the diameter by 3.1416. 

Sphere: Cubic contents=.5236D3 

Superficial area=3.1416D2 
The area of the surface of a sphere is equal to 
the area of the surface of a cylinder, the diameter 





U- D ^ 

Fig. 99. 

and the height of which are each equal to the 
diameter of the sphere. Also, the area of the sur- 
face of a sphere is equal to four times the area of 
its diameter. 

The latter definition is easily remembered, and 
is useful in calculating the areas, of the hemi- 
spheres, because the area of the sheet or disc of 
metal required for raising a hemisphere must be 



40 



MECHANICAL DRAWING 



equal in area to the combined areas of two discs, 
each equal tO' the diameter of the hemisphere. 

To find the cubic contents of a hemisphere — 
Fig. 99. Multiply the cube of the diameter by 
.2618. 

To find the superficial area of a hemisphere. 
Hemisphere: Cubic contents=.2618D2 
Superficial area^2.3562D^ 

To find the cubic contents of a cylindrical ring — 
Fig. 100. To the cross-sectional diameter of the 




ring add the inner diameter of the ring, multiply 
the sum by the square of the cross-sectional diam- 
eter of the ring and by 2.4674, the product is the 
cubic contents. 

To find the superficial area of a cylindrical ring. 
To the cross-sectional diameter of the ring add the 
inner diameter of the ring. Multiply the sum by 
the cross-sectional diameter of the ring and by 
9.8696, the product is the superficial area- 



Cylindrical ring: Cubic contents=2.4674T^ (T+ 
H) 

Superficial area^9.8696T (T+ 

H) 
D=(H+2T) 

To find the cubic contents of a cylinder — Fig. 
101. Multiply the area, of one end by the length 
of the cylinder, the product will be the cubic con- 
tents of the cylinder. 

To find the superficial area of a cylinder. Mul- 





tiply the circumference of one end by the length 
of the cylinder and add to the product the area 
of both ends. 
Cylinder: Cubic contents= .7854 (D+H) 

Superficial area^l.5708D (2H+D) 
To find the cubic contents of a cone — Fig. 102. 
Multiply the square of the base by the perpendic- 
ular height and by .2618. 



MENSURATION 



41 



To find the superficial area of a cone. Multiply 
the circumference of the base by one-half the slant 
height and add to the product the area of the base. 

Cone: Cubic contents=.2618 (D^xH) 
Superficial area^.7854 (2S+D) 

To find the cubic contents of the frustum of a 
cone — Fig. 103. To the sum of the areas of the 
two ends of the frustum, add the square root of 
the product of the diameters of the two ends, this 
result multiplied by one-third of the perpendicular 
height of the frustum will give the cubic contents. 

To find the superficial area of the surface of the 
frustum of a cone. Multiply the sum of the diam- 
eters of the ends by 3.1416 and by half the slant 
height. Add to the result the area of both ends 
and the sum of the two will be superficial area. 
Frustum of cone: 



Cubic Contents 



H(.2618 (E2-fD2)v'EXD) 



(D-l-E\ 
— ^)+.7854(E2-f-D2) 



-V(^^ 



-fH2 



To find the contents of a cube — Fig. 104. The 

contents of equal to the cube of its diameter. 

To find the superficial area of a cube. The su- 
perficial area of a cube is equal to six times the 
square of its diameter. 



Cube: Ctibic contents=D^ 

Superficial area^GD^ 

To find the cubic contents of a rectangular solid 





-^ A 


j«~E— *j 


f 


U — p — ->i 

Fig. 103. 


< D »-' 

Fig. 104. 



— Fig. 105. Multiplying together the length, 
width and height will give the cubic contents of 
the rectangular solid. 




Fig. 105. 

To find the superficial area of a rectangular 
solid. Multiply the width by the sum of the 
height and length and add to it the product of the 



42 



MECHANICAL DRAWING 



height multiplied by the length, twice this sum 
is the superficial area of the rectangular solid. 
Rectangular solid: 

Cubic contents=DxHxL 

Superficial area^2 (D (H+L)+HL) 

To find the cubic contents of a pyramid — Fig. 

106. Multiply the area of the base by one-third 

the perpendicular height and the product will be 

the cubic contents of the pyramid. 

To find the superficial area of a pjn^amid. Mul- 




Vc D >i 

Fig. 106. 



tiply the circumference of the base by half the 
slant height and to this add the area of the base, 
the sum will be the superficial area. 



Pyramid: Cubic contents = 



D^XH 



n . , /4D + S , , \ 

Superficial area = ( — ^ h4D I 



Hf 



-f-H2 



MENSURATION OF TRIANGLES. 

To find the base of a right-angle triangle when 
the perpendicular and the hypothenuse are given 
— Fig. 107. Subtract the square of the perpendic- 
ular from the square of the hypothenuse, the 
square root of the difference is equal to the length 
of the base. 



Base=T/Hypothenu8e2 — Perpendicular^ or B=i/C^ — H^ 

To find the perpendicular of a right-angle tri- 




rig. 107. 

angle when the base and hypothenuse are given. 

Subtract the square of the base from the square 
of the hypothenuse, the square root of the differ- 
ence is equal to the length of the perpendicular. 



Perpendicular =y'Hypothenuse^ — Base^ or H=v/C^ — B^ 

To find the hypothenuse of a right-angle tri- 
angle when the base and the perpendicular are 
^ven. The square root of the sum of the squares 



MENSURATION 



43 



of the base and tlie perpendicular is equal to the 
length of the hypothenuse. 

Hypothenuse=i/Base^+Perpendicular^ 



C=i/B2+H2 

To find the perpendicular height of any oblique 
angled triangle — Fig. 108. From half the sum 
of the three sides of the triangle, subtract each 
side severally. Multiply the half sum and the 
three remainders together and twice the square 







^ A 


\ \ 






^ / 1 


\ \ 


'' / 


\ \ 


'' / 


\ \ 


y / 1 






1 



Us- c *l 

Fig. 108. 

root of the result divided by the base of the tri- 
angle will be the height of the perpendicular. 



D= 



2i/S(S— A) (S— B) (S— C) 



s=- 



C 

Sum of sides 



To find the area of any oblique angled triangle 
when only the three sides are given. From half 
the sum of the three sides, subtract each side sev- 
erally. Multiply the half sum and the three 
remainders together and the square root of the 
products is equal to the area required. 



Area=-i/S(S— A) (S— B) (S— C) 

To find the height of the perpendicular and the 
two sides of any triangle inscribed in a semi-circle, 





^A 




^^'^ ^ 


'*^.-^ 


^^ ^-"^ 


r> 


K "^ 


/ JO^^^ 


/I 
1 


NN> 







W^x 


/x' ^^^ 


1 


\ \ ^ 


^\ ^^ 


1 


i 


\ ' 







k A -*U-B-»J 

Fig. 109. 

when the base of the triangle and the location of 
the perpendicular are given — Fig. 109. 



^ B 



B= 



C2 



C=i/AXB 



D=t/A(A+B) E=i/B(A-|-B) 



THE MECHANICAL POWERS 



Mechanical Powers consist of simple meclianical 
devices whereby weights may be raised or resist- 
ances overcome with the exertion of less power 
than would be necessary without them. 

They are six in number: The lever, the wheel 
and pinion, the pulley, the inclined plane, the 
wedge, and the screw. Properly two of these com- 
prise the whole, namely, the lever and the inclined 
plane, — the wheel and pinion being only a lever of 
the first kind, and the pulley a lever of the sec- 
ond, the wedge and screw being also' similarly 
allied to that of the inclined plane. Although such 
seems to be the case, yet they each require, on 
account of their various modifications, a different 
rule of calculation adapted expressly to the differ- 
ent circumstances in which they are required to 
act. 

The primary elements of machinery are there'- 
f ore two only in number, the lever and the inclined 
plane. 

The Lever. 

Levers, according to the method of application, 
are of the first, second, or third kind. Although 
levers of equal lengths produce different effects, 
the general principles of estimation in all are the 



same, namely, the power is to the weight, as the 
distance of one end of the fulcrum is to the dis- 
tance of the other end to the same point. 

In a lever of the first kind the fulcrum is be- 
tween the power and 



the weight, as in Fig. 
110. A pair of pliers 
or scissors are double 
levers of the first 
kind. 

In a lever of the 
second kind, the 
weight is between the 
power and the ful- 
crum, as in Fig. 111. 
A wheel-barrow, or 
the oars of a boat 
where the water is con- 
sidered the fulcrum, 
and a door, represent 
levers of the second 
kind. 

In a lever of the 
third kind, the power 
is between the fulcrum 
and the weight, as in 



w 



11 



Fig. 110. 




W F 



TT 



Fig. 111. 




w 



F P 



Fig. 112. 



44 



THE MECHANICAL POWERS 



45 



Fig. 112. Levers of the third kind are instru- 
ments such as tongs, shears, etc. 

In the first kind, the power is to the weight, as 
the distance W F is to the distance F P. 

In the second, the power is to the weight, as the 
distance F W is to that of F P; and. 

In the third, the weight is to the power, as the 
distance F P is to that of F W. 

To find the power. Multiply the weight by its 
distance from the fulcrum, and divide by the dis- 
tance of the weight from the fulcrum. 

To find the weight. Multiply the power by its 
distance from the fulcrum, and divide by the dis- 
tance of the weight from the fulcrum. 

To find the distance of the power from the 
fulcrum. Multiply the weight by its distance from 
the fulcrum, and divide by the power. 

To find the distance of the weight from the 
fulcrum. Multiply the power by its distance from 
the fulcrum, and divide by the weight. 

Let P be the power, F the fulcrum and W the 
weight, then for a. lever of the first kind. (Fig. 
110.) 



P=W 



FW 
FP 



W=P 



FP 

FW 



And for a lever of the second kind. (Fig. 111.) 



P=W 



FW 
FP 



W=P 



FP 
FW 



And for a lever of the third kind. (Fig. 112.) 



P=W 



FW 
FP 



W=P 



FP 
FW 



The Wheel and Pinion. 

The mechanical advantage of the wheel and 
pinion system. Fig. 113, is as the velocity of the 
weight to the velocity of the power, and being only 
a modification of the first kind of lever, it of 
course partakes of the same principles. 




i*D-»l< — R — *l 

® 

Fig. 113. 

To find the power. Multiply the weight by the 
radius of the drum, and divide by the radius of the 

wheel. 

To find the radius of the wheel. Multiply the 
weight by the radius of the drum, and divide by 
the power. 



46 



MECHANICAL DRAWING 



To find the radius of the drum. Multiply the 
power by the radius of the wheel, and divide by 
the weight. 

To find the weight. Multiply the power by the 
radius of the wheel, and divide by the radius of 
the drum. 

Let W be the weight, D the radius of the drum, 
R the radius of the wheel and P the power re- 
quired to lift the weight, then for a Wheel and 
Drum system: (Fig. 113.) 



WXD 

R 



PINION 



D= 



PXR 



W 



E= 



WXD 



W= 



PXR 



D 




W= 



.WXDXP 

RXG 
PXRXG 



DXP 



For a Crank, Pinion and Gear and Drum sys- 
tem: (Fig. 114.) 

To determine the amount of effective power 
produced from a given power by means of a crank, 
pinion and gear, and drum system. Multiply the 



diameter Of the circle described by the crank or 
turning handle by the number of revolutions of 
the pinion to one of the wheel. Divide the prod- 
uct by the diameter of the drum and the quotient 
is the ratio of the effective power to the exertive 
force. Fig. 114. 

Given any two parts of a crank, pinion and gear, 
and drum system, to find the third, that shall pro- 
duce any required proportion of mechanical effect. 
Multiply the two given parts together, and divide 
the product by the required proportion of effect, 
the quotient is the dimensions of the other part. 



.PXRXG 
WXP 



R= 



WXDXP 



PXG 



iP=Either pitch diameter or number of teeth in 

the pinion. 
JG}==Either pitch diameter or number of teeth in 

the gear. 
Let E be the ratio of the effective power to the 
effective force produced, then 

.RXG 



E=3.1416 



DXP 



The Pulley or Sheave. 

The pulley or sheave is a wheel over which a 
rope is passed to transmit the force applied to the 
cord in another direction. There are two kinds 
of pulleys, the one turning on fixed centers, the 
other tumrag on traversing centers. 



THE MECHANICAL POWERS 



4? 



The fixed or stationary pulley (Fig. 115). This 
acts like a lever of the first kind. It affords no 
mechanical advantage, and merely changes the 
directidfl of the force, and does not alter its in- 
tensity, but it affords great facilities in the appli- 
cation of force, as it is easier to pnll downwards 
than upwards. In this class of pulley the power 
is equal to the weight to be raised. 



//M/M/Mm////m^^^^ 




r> 




The movable pulley (Fig. 116). This acts like 
a lever of the second kind. One end of the rope 
is suspended to a fixed point, as a fulcrum, in a 
beam, and the weight is. attached to the axis of 
the pulley. This kind of pulley doubles the power 
at the expense of the speed, and the product of the 
power by the diameter of the pulley, is equal to 



the product of the weight by the radius of the 
pulley. 

A movable pulley acting as a lever of the third 
kind is shown at Fig. 117. One end of the cord 
is fixed to a floor, and the weight is attached to 
the other end, the power being applied to the 
axis. The power is equal to twice the weight, and 
the product of the power by the radius of the pul- 




Fig. 117. 




Fig. 118. 



ley is equal to the product of the weight by the 
diameter of pulley. In the arrangement shown at 
Fig. 118 the power is equal to one-half the weight. 
A combination of movable pulleys with separate 
and parallel cords is shown at Fig. 119. Each 
system reduces the resistance to the extent of 
one-half, hence the power may be found by divid- 
ing and subdividing the weight successively by 



48 



MECHANICAL DEAWING 



To find the power. 



2, as many times as there are movable pulleys. 

The weight may be found by multiplying the 

power successively by 2, as many times as there 

are movable pulleys. 

Divide the weight to be 
raised by the number of 
cords leading to, from, 
or attacked, to the power 
block. The quotient is 
the power required to 
produce an equilibrium, 
provided friction did not 
exist. When the fixed end 
of the rope is attached 
to the fixed block, the 
power may be found by 
dividing the weight by 
twice the number of 
movable pulleys. When 
the fixed end of the rope 

®is attached to the mov- 
able block, the power 
Pj -^-j^g may be found by divid- 

ing the weight by twice 
the number of movable pulleys plus 1. 

To find the number of sheaves or pulleys re- 
quired. Divide the power to be raised by the 
power to be applied; the quotient is the number 
of sheaves in, or cords attached to the rising 
block. 




To find the weight that will be balanced by a 
given power. When the rope is attached to the 
fixed block, multiply the power by twice the num- 
ber of movable pulleys. 

When the rope is attached to the movable block 
multiply the power by twice the number of mova- 
ble pulleys plus 1. 

The Inclined Plane. 

The inclined plane (Fig. 120) is properly the 
second elementary power, and may be defined the 
lifting of a load by regular instalments. In prin- 




ciple it consists of any right line not coinciding 
with, but laying in a sloping direction to, that 
of the horizon, the standard of comparison of 
which commonly consists in referring the rise to 
so many parts in a certain length or distance, as 
1 in 100, 1 in 200, etc., the first number represent- 
ing the perpendicular height, and the latter the 
horizontal length in attaining such height, both 
numbers being of the same denomination, unless 
otherwise expressed. 



TTHE MECHANICAL POWERS 



49 



In using an inclined plane for the purpose of 
raising loads to a higher level, the power is ap- 
plied parallel tO' the inclined plane, and the weight 
is raised in opposition to gravity, the work done 
on it is expressed by the product of the weight 
and the vertical height of the inclined plane. 

The advantage gained by the inclined plane, 
when the power acts in a parallel direction to the 
plane, is as the length to the height. 

To find the power. Multiply the weight by the 
height of the plane, and divide by the slant length. 
The quotient is the power. 

To find the weight. Multiply the power by the 
slant length of the plane, and divide by the 
height. 

To find the height of the inclined plane. Mul- 
tiply the power by the slant length, and divide 
by the weight. 

To find the slant length of the inclined plane. 

Multiply the weight by the height of the plane, 
and divide by the power. 

Let W be the weight to be drawn up the in- 
clined plane, H the height and S the slant length 
of the incline. If P be the power required to draw 
the weight W up the inclined plane, then 



WXH 

S 



w= 



PXS 
H 



H= 



PXS 
W 



The Wedge. 

The wedge is a double inclined plane, conse- 
quently its principles are the same. When two 
bodies are forced asunder by means of the wedge 
in a direction parallel to its head: Multiply the 
resisting power by half the thickness of the head 
or back of the wedge, and divide the product by 



!*H*i 




Fig. 121. 

the length of one of its slant sides. The quotient 
is the force required equal to the resistance. 

F=Force required. P=Resisting power. 



F= 



PXH 



25 



P= 



FX25 
H 



(Fig. 121.) 



S=-/L2+H2 



S=Slant side of wedge=-v/-T — hi? 

When only one of the bodies is movable, the 
whole breadth of the wedge is taken for the mul- 
tiplier, and the following rules are for such 
wedges, acting under pressure only on the head 
of the wedge, or at the point of the wedge by 
drawing. 



50 



MECHANICAL DEAWING 



To find the transverse resistance to the wedge 
or weight. Multiply the power by the length of 
the slant side of the wedge, and divide by the 
breadth of the head. 

To find the power. Multiply the weight or 
transverse resistance by the breadth of the head 
and divide by the length of the slant side of the 
wedge. 

To find the length of the slant side of the wedge. 




Multiply the weight by the breadth of the wedge 
and divide by the power. 

To find the breadth of the wedge. Multiply the 
power by the length of the slant side of the wedge, 
and divide by the weight. 

F= Force required. P=Kesisting power. 



s 



H 



S=Slant side of wedge=VB.^+h^ 

Note. — ^For all practical purposes the length L 
may be used instead of the slant S of the side. 



The Screw. 

The screw, Fig. 123, in principle, is that of an 
inclined plane wound around a cylinder which 
generates a. spiral of uniform inclination, each 
revolution producing a rise or traverse motion 
equal to the pitch of the screw, or distance be- 
tween two consecutive threads. The pitch being 
the height or angle of inclination, and the circum- 




__R J 



Fig. 123. 

ference the length of the plane when a lever is not 
applied. The lever being a necessary qualification 
of the screw, the circle which it describes is taken, 
instead of the screw 's circumference, as the length 
of the plane, the mechanical advantage is there- 
fore as the circumference of the circle described 
by the lever where the power acts, is to the pitch 
of the screw, so is the force to the resistance. 
As the circumference of a circle is equal to the 



THE MECHANICAL POWERS 



51 



radius multiplied by twice 3.1416, or 6.2832, hence 
the following rules for the screw. 

To find the power. Multiply the weight by the 
pitch of the screw and divide by the product of 
the radius of the handle by 6.2832. 

To find the weight. Multiply the power by the 
product of the radius of the handle by 6.2832 and' 
divide by the pitch of the screw. 

To find the pitch of the screw. Multiply the 
power by the product of the radius of the handle 
by 6.2832 and divide by the weight. 



To find the length or radius of the handle. 

Multiply the weight by the pitch of the screw 
and divide by the product of the power by 6.2832. 

P=Lifting power of jack. R=Length of lever. 
F=Force required at end of lever. 
N=Nuniber of threads per inch of jack screw. 



P=6.283(NXEXF) 
P 



F= 



N= 



6.283 (RXF) 



R= 



6.283 (NXR) 

P 
'6.283(NXF) 



THE DEVELOPMENT OF CURVES 



The Spiral is a curve generated by a point mov- 
ing in a plane about a center from which its dis- 
tance is continually increasing. 

Imagine a right line, AB, Fig. . 124, free to 
revolve in the plane of the paper about one of its 




Fig. 124 — The Spiral of Archimedes. 

extremities. A, as an axis. Also, conceive a point 
free to move on this line. Three classes of lines 
may now be derived in the following manner: If 
the line be stationary and the point moves, a 



straight line will be generated. If the point be 
stationary and the line revolves, a circle will be 
generated. If both point and line move, a spiral 
will be generated. By varying the relative motion 
of point and line the character of the curve will be 
changed and the various classes of spirals 
described. 

The Spiral of Archimedes — Fig. 124. If the mo^ 
tion of the line and the point be uniform the 
equable spiral will be generated. The line AB is 
called the kadius vector, and the radial distance 
traversed by the point during one revolution of 
the radius vector is called the pitch. Twelve suc- 
cessive positions of the radius vector are shown by 
AC, AD, AE, etc. The distance of the point from 
the center being increased by one-twelfth of the 
pitch, AC, for each one-twelfth of a revolution of 
the radius vector. In practice, determine at least 
twenty-four points, and lightly sketch the curve 
freehand. 

The Logarithmic Spiral. The- construction of 
this curve is based on the principle that any radius 
vector, as AD, Fig. 125, which bisects the angle 
between two other radii, as AC and AB, is a mean 



52 



th:e development of curves 



53 



proportional between them, that is, (AD) 2= 
AOxAB. This spiral is called equiangular he- 
cause the angle between any radius vector and the 
tangent to the curve at its extremity is constant. 

If B and C are points in the spiral and the ratio 
of AO and AB be given, the intermediate point D 
may be obtained by describing a semicircle on BC 
as a diameter and erecting a perpendicular at A. 




Fig. 125 — The Logarithmic Spiral. 

Its intersection with the semicircle at D will deter- 
mine the required point, and AD will be a mean 
proportional between AC and AB. Other points 
on the cnrVe lying on the diameters BC and DF 
may b^ obtained by intefsecting these diameters 
with the pefpendiculaf s CL, LM, MN, etc. Again, 
having points C and D, bisect DAC and determine 
a mean proportional between AC and AD. This 



may be done by laying off AF equal to AC and 
determining the mean proportional AK as before. 
Then lay off AE equal to AX. Other points on the 
diameters ER and T may be obtained by perpen- 
diculars as indicated. 

The Ellipse is a curve generated by a point mov- 
ing in a plane so that the sum of the distances 




Fig. 126— The Ellipse— 1. 

from this point to two fixed points shall be con- 
stant. Fig. 126, if EKF be a cord fastened at its 
extremities, E and F, and held taut by a pencil- 
point at K, it may be seen that as motion is given 
to the point it will be constrained to move in a 
fixed path dependent on the length of the cord. 
When the pencil-point is at B, one segment of the 
cord will equal BE and the other BF, their sum 



54 



MECHANICAL DRAWING 



being the same as KE plus KF, and also equal to 
AB. Tlie fixed points E and F are called the foci. 
They lie on the longest line that can be drawn 
terminating in the curve of the ellipse. The line 
is known as the major axis, and the perpendicular 
to it at its middle point, also terminating in the 
ellipse, is the minor axis. Their intersection is 
called the center of the ellipse, and lines drawn 
through this point and terminating in the ellipse 
are known as diameters. When two such diam- 
eters are so related that a tangent to the ellipse 
at the extremity of one is parallel to the second, 
they are called conjugate diameters. KL and MN 
are two such diameters. 

In order to construct an ellipse it is generally 
necessary that either of the following be given: 
The major and minor axes, either axis and the foci, 
two conjugate diameters, a chord and axis. 

A series of points must be chosen that the sum 
of the distances from either of them to the foci 
must equal the major axis. Thus, HE+HF must 
equal CE+GF, or KF+KE, each being equal to 
AB. If the major axis and the foci be given to 
draw the curve, points may be determined as fol- 
lows: From E, with any radius greater than AE 
and less than EB, describe an arc. From F, with 
a radius equal to the difference between the major 
axis and the first radius, describe a second arc 
cutting the first. The points of intersection of 
these arcs will be points, the sum of whose disp 



tances from the foci will equal the major axis, and 
therefore points of an ellipse. Similarly find as 
many points as may be necessary to enable the 
curve to be drawn free-hand. 

Having given the major and minor axes, find 
the foci by describing, from C as a center, an arc 
with a radius equal to one-half the major axis. 
The points of intersection with the major axis will 




The Ellipse— 2. 



be the foci, and this must be so since the sum of 
these distances is equal to' the major axis, and the 
point C being midway between A and B the two 
lines CE and CF must be equal. Again, if the 
major axis and foci are given, with a radius equal 
to one-half this axis describe arcs from the foci 
cutting the perpendicular drawn at the middle 
point of the major axis and thus obtain the minor 
axis. Having the two axes proceed as before. 



THE DEVELOPMENT OF CURVES 



55 



A tangent to an ellipse may be drawn at any 
point, K, by producing FK and dissecting the 
angle SKEi, the bisecting line, KT, will be the re'- 
quired tangent. 

Having given the major and minor axes in Fig. 
127. Ftom the extremity of the major axis, draw 
B6 parallel and equal to the minor axis, and divide 
it into any number of equal parts, in this case six. 
Divide BGr into the same number of equal parts. 




Fig. 128— The Ellipse— 3. 

Through points 1, 2, 3, etc., on B6, draw lines to 
extremity C of the minor axis. From D, the other 
extremity of the minor axis, draw lines through 
points 1, 2, 3, etc., on BG, intersecting the above 
lines in points which will lie in the re- 
quired ellipse. Construct the remainder of the 
ellipse in the same manner. 



Having given an axis CD and chord FH as in 
Fig. 128. From F draw F4 parallel to CD, divide 
it into any number of equal parts, in this case 
four. Divide the half chord FE into' the same 
number of equal parts, through these points and 
extremities of given axis draw intersecting lines 
as before, thereby obtaining the elliptical arc FD. 
Construct opposite side in the same manner. 

The Parabola is a curve generated by a point 
moving in a plane so that its distance from a 




Fig. 129— The Parabola. 

fixed point shall be constantly equal to its dis- 
tance from a given straight line. Point F in Fig. 
129 is the focus, CD is the given straight line 
called the directrix, and AB, a perpendicular to 
CD through F, is the Axis V, the intersection of 
the axis with the curve, is the vertex, and it must 
be equidistant from the focus and the directrix, 



56 



MECHANICAL DRAWING 



Having given the focus F and the directrix CD. 
Bisect FA to find the vertex V. Through any point 
on the axis, as L, draw MN parallel to the directrix 
and with radius LA describe arc 1 from focus F 



1 2 3 4 6 6 7 




Fig. 130. 

as center, intersecting line MN at points M and N. 
These are points in the parabola. Similarly obtain 
other points and draw the required curve. A tan- 



gent to the curve may be drawn at any point M by 
drawing MO parallel to the axis and bisecting the 
angle OMF. MT' is the required tangent at the 
point M. 

Since the angle T'ME. is equal to the angle TMF, 
it follows that MR would be the direction of a ray 
of light emitted from the focus F and reflected 
from the parabola at M. 

Having given the abscissa VB, and the ordinate 
BE in Fig. 130. Draw AE parallel and equal to VB. 
Divide AE and BE into the same number of equal 
parts. From the divisions on BE draw parallels 
to the axis and from the divisions on AE draw 
lines converging to the vertex V. The intersection 
of these lines, 1 and 1, 2 and 2, etc., will determine 
points in the required curve. In like manner ob- 
tain the opposite side. 



THE DEVELOPMENT OF SURFACES. 



To develop a square tapering article. The plan 
and vertical height or elevation axe shown in Fig. 
131. Draw the diagonals and take the distance 




from the center a to b, and mark off the same from 
g to d. Take the distance from a to 1 or k, and 
mark off the same from h to e. Draw a line through 



the points d, e, to cut the perpendicular line at f. 
Then draw the perpendicular line af , Fig. 132, and 
take the radius fd, Fig. 131, and with it describe 
the arc of a circle hdk, Fig. 132. With the radius 




57 



Fig. 132. 

fe in Fig. 131, and with f in Fig. 132 as a center, 
draw the smaller arc e. Take the length of one 
side of the base from c to b, Fig. 131, and mark off 
the same four times on the circle hdk at h, g, d, i, k. 



58 



MECHANICAL DRAWING 



Draw through these points to the center f, join 
these points hg, gd, di, and ik. Also join the points 
on the smaller circle in the same manner, which 
will complete the development. 



Fig. 133. 




Ta develop a rectangular taper-sided tray. The 

vertical height and one-half the plan are shown in 
Fig. 133. Draw the horizontal line bd and the per- 
pendicular line op as in Fig. 134. Draw the rec- 
tangle efgh the same size as efgh in Fig. 133. Take 
the length ab as in Fig. 133 and mark off a corre^ 
spending distance from e to b, h to d, and to p, 
as in Fig. 134, and draw through the points b, p, 




Fig. 134. 



Fig. 136. 



and d the lines at right angles as bq, st, and dr. 
Transfer the leng-th il to bq and tO' dr, also the 
length ul from p to s, and from p to t. Then draw 
the lines qf, sf, tg, and rg, which will complete one- 
half of the development. 

To develop a hexagon tray with tapering sides. 
The elevation and one-half the plan are shown in 



THE DEVELOPMENT OF SURFACES 



59 



Fig. 135. To" develop the pattern draw the perpen- 
dicular be, Fig. 136, and draw the half-hexagon 
efghi, of the same size as efghi in Fig. 135. Divide 
the lines hg and gf intO' two equal parts and draw 
the lines ak and am through the points of bisec- 

a 




Fig. 137. 

tion, then carry the length ab, Fig. 135, from 1 to 
k in Fig, 136. Draw through k the line no parallel 
tO' hg. Then take the kl, Fig. 135, and mark off the 
same from k to^ n and from k to o, and draw the 



lines hn and go. Proceed in the same manner to 
draw the remainder, which will complete one-half 
of the development. 

To develop an oblique pyramid. The lengths of 
the sides are shown projected, a'b to Cb', and a'a 
to Cc/ giving for the true lengths a'b' and a'c'. 




Fig. 138, 



Take the length a'c' in Fig. 137, and with it strike 
the radius a'c in Fig. 138. With the length a'b' in 
Fig. 137, strike the radius a'b in Fig. 138. Take 
the length of one side as ba in Fig. 137, and set it 
off from e to a, and from a to c, from c to b, and 



60 



MECHANICAL DRAWING 



from b to e. Connect the points of intersections of 
tlie arcs by means of straight lines as ae, ac, cb, 
and be. Also a' with e and e, and the outline will 
be described. 



a 

,4 
/! 




1 i 1 






^^-^"^^ ^\^ 



:^a' 



Fig. 139. 

To develop an oblique truncated pyramid. The 

correct lengths of the sides are shown projected 
in Fig. 139 as in the previous figure. The leng-ths 
ab", ac" on the plane D, are also the correct 
lengths for the sides of the small end of the pyra- 



mid. In Fig. 140 the outline of the base is devel- 
oped precisely the same as in the previous exam- 
ple. To develop the top edge, the lengths c'c" in 
Fig. 139 are transferred to ac", cc" in Fig. 140, and 
the lengths b'b" in Fig. 139 to b'b", eb" in Fig. 140. 




Connecting these points with straight lines gives 
the outline of the development. 

To develop a cone cut in elliptical section. Fig. 
141 shows the cone at b-b, the cut section forming 
or having the shape of an ellipse and Fig. 142 is 
the development of the lower part of the cone. Let 



THE DEVELOPMENT OF SURFACES 



61 



ABC, Fig. 141, represent the outline of the cone. 
Strike a semicircle BGO, equal in radius to half 
the length of the base BO, and divide it into any 

A 



A' \h b 




BA( 



'^-^1^-^H 



-4 

G 

Fig. 141. 

number of equal parts as B, D,. E, F, G, H, I, J, 0. 

Carry perpendicular lines up to cut the line BO in 




n 



rig. 142. 



62 



MECHANICAL DRAWING 



d, e, f, g, h, i, j, and draw lines from these points 
to the apex at A. They will cut the diagonal b-b 
at d'e'f'g'h'i'j', then carry horizontal lines from 
these points to' meet the slant edge B'A in d", e", 
f'', g", h", i^', y\ Then the lengths Bd", Be'', Bf", 
feg", Bh", Bi", Bj", will be the actual lengths of 
the lines dd', ee', ff gg\ hh', ii", jj" in Fig. 142. 
To describe the pattern take the length AB as a 
radius and strike an arc of a circle as ABCD. 
From the point set off to the right and left the 
points J, I, H, G, F, £, D, B, using the lengths of 



the points of division in Fig. 141. Draw lines from 
the points of division to A. On these lines set off 
the projected lengths from Fig. 141, thus: 
Take the length CC and set off from C to C 
in Fig. 142. Take the length from B to j' and 
set it off from j to j' in Fig. 142. Take the 
length B to i'' and set it off from i to i', and so 
on. A curve drawn through the points c', j', i', h', 
g', r, e', and d' to right and left will give the 
shape of the development. 



THE INTERSECTION OF SURFACES 



To develop the intersection of a round elbow at 
right angles. Draw ABCFED, which is the size of 
the elbow required. On the line CF, Fig. 143, 
strike a semicircle of the same diameter as the 
pipe. Divide the semicircle into any number of 



F on each side, and draw the perpendicular line 
Fn, em, dl, ck, bi, ah, Cg. Extend the line BO 
to cut the perpendicular Cg, and draw lines from 
the points a, b, c, d, and e in the semicircle to cut 
the perpendiculars at h, i, k, 1, and m. Draw a 




equal parts as a, b, c, d, e. Draw the line FF in 
Fig. 144, and make it equal to twice the length of 
the circumference of the semicircle in Fig. 143, 
by setting the parts a, b, c, d, e from C to F and 



63 



curve through all the points of intersection, as 
n, m, 1, k, i, h, and g. This will form the curve 
for half of the development. 
To develop the intersection of two pipes which 



64 



MECHANICAL DRAWING 



intersect at an angle. Let DAEC represent the 
larger pipe, and let HFJG be drawn tO' the re^ 
quired size of the pipe that is to be connected with 
it at any desired angle. Draw the line FG, Fig. 




X — 1 — 


1 ' 1 


1 
1 



Fig. 147. 



Fig. 146. 



145, at right angles with FH, on FG describe a 
semicircle, and divide into any number of equal 
parts, as 1, 2, 3, 4, 5, draw lines through these 
points at right angles with FH. Then describe a 



semicircle ABC representing the diameter of the 
larger pipe, and extend the line DA to M and CA 
to' L, take the radius OF of the smaller semicircle 
and from A mark off the same distance to M, then 
take the half length of the base of the smaller 
pipe as K to H or H to J and mark off a like dis- 
tance from A to L. A quarter of an ellipse is 
required as shown from L to M, the radius of 
which may be obtained in the following manner: 
Draw a line from M tO' N, also one from L to N at 
right angles, and draw the diagonal line LM; 
draw a line from the point N to cut the diagonal 
LM at right angles, producing the points c and b. 
With c as a center and radius cL draw a curve 
from L to a. With b as a center and radius ba, 
draw the remainder of the curve from a to M. 
Divide the curve from L to M into three equal 
parts, and then draw perpendicular lines from 
these points tO' intersect the semicircle ABO, as 
df, eg, Mh. Draw lines parallel to AD from f to 
i, g to k, and h tO' 1. The points where these lines 
are intersected by the lines drawn from the semi- 
circle on the smaller cylinder will be the points 
through which to draw the curve r, s, t, w, v. 
Draw the line GG, Fig. 146, equal to the circumfer- 
ence of the smaller pipe, or to twice the number 
of divisions in the small semicircle. Divide one- 
half of GG into six equal parts, as Fm, mn, no, op, 
pq, and qG. Draw perpendicular lines from F, m, 
n, 0, p, q, and G. Take the length of the lines in 



THE INTERSECTION OF SURFACES 



65 



Fig. 145, as FH, mr, ns, ot, pu, qv, and GJ, and 

transfer their lengths to the perpendicular lines 
marked by corresponding letters. Draw a curve 
through the points thus obtained as H, r, s, t, w, 
V, and J. This will give the half -pattern for the 




Fig. 148. 



smaller pipe. To obtain the curve for the hole in 
the large pipe. Draw DB and HJ, Fig. 147, at 
right angles, take the distances in Fig. 145, from 
A to f, g and h, and mark off like distances on each 



side of A in Fig. 147 on the line DB, as f, g, and h, 
and draw lines from these points parallel to HJ. 
Draw a perpendicular line from the point K to R 
in Fig. 145, and transfer the lengths KH and KJ, 
from A to H and A to J in Fig. 147, also the dis- 
tances xr and xv from f to r and f to v in Fig. 147, 
and the distances y to s and y to u, from g to s 






C 8 



5 4 
Fig. 149. 



1 A 



and g to u. Take the distance from R to t in Fig. 
145 and mark off from h to t in Fig. 147. The 
curve drawn through these points will give the 
shape of the opening in the larger pipe. 

To develop an intersection for a joint at any 
angle in a circular pipe. Let ABO, Fig. 148, be 
the half diameter of the pipe. Draw the line AC 
and the lines EG and DH, then draw the line ED, 



66 



MECHANICAL DRAWING 



cutting the lines at tlie points E and D. Divide 
the circumference of the semicircle ABC into any 
number of equal parts, and from these points draw 
lines parallel to AF, as 1, 2, 3, 4, 5, 6, 7, and 8. 
Then lay off the line AC, Fig. 149, equal in length 
to the circumference of the semicircle ABC. Erect 
the lines AE and CD at right angles to AC, and 
then lay off on the line AC the same number of 



spaces as on the circumference of the semicircle 
ABC, and from these points draw lines parallel to 
AE, as 1, 2, 3, 4, 5, 6, 7, and 8. Make AE equal to 
AE in Fig. 148 and CD to CD in the same figure, 
also each of the parallel lines bearing the numbers 
1, 2, 3, 4, 5, 6, 7, and 8. A curve drawn through 
these points will give the shape of the half inter- 
section. 



GENERAL INSTRUCTIONS FOR MACHINE DRAWING 



Line Shading. When it is necessary to sug- 
gest the character of a surface without a second 
view, line shading may be used. Fig. 150 illus- 
trates a good method for the shading of a cylindri- 
cal surface. The lines may be equally spaced, al- 
though the appearance is somewhat improved by 





Fig. 150. 



Fig. 151. 



increasing the space as the center line is ap- 
proached, and decreasing the space on the lower 
half, while increasing the width of the line. In 
shading a concave surface, as in Fig. 151, the oper- 
ation is reversed. Spherical, and other classes of 
surfaces may be represented by this method as in 
Fig. 152, but the rendering of them requires con- 



siderable skill, and it is beyond the scope of this 
work to consider the various methods. 

Sections. It is frequently necessary to make 
the representation of an object as it would appear 
if cut by a plane, and with the portion nearer the 
observer removed. This cut section is indicated 
by a series of parallel lines, usually drawn at an 




67 



Fig. 152. 

angle of 45°. The width of the lines is the same 
as a fine line, and the distance between the lines 
is determined by the area and intricacy of the 
section. These lines are not to be drawn in pen- 
cil, but only in ink. An instrument known as a 
section liner is designed to insure accuracy in the 
spacing, but the draftsman should learn to judge 
this by the eye. In doing so, care must be used 



68 



MECHANICAL DEAWING 



to avoid making too small a space, and it is desir- 
able to try the spacing on a separate slieet before 
sectioning the drawing. A greater degree of uni- 
formity may be obtained by looking back after 
drawing every three or four lines. 

The different surfaces in the plane of the sec- 
tion are indicated by changing the direction of the 



Section Lining or Brass Hatching Shown in 


Fig. 153. 


A 


B 


C 


D 


E 


F 


Cast Iron. 


W. Iron. 


e. steel. 


W. Steel. 


T. Steel. 


Brass. 


G 


H 


1 


J 


K 


L 


Copper. 


Lead 

or 

Babbitt. 


Aluminum. 


Rubber 

or 

Vulcanite. 


Leather. 


Asbestos. 


M 


N 


o 


P 


Q 


R 


Wood. 


Brick. 


stone. 


Glass., 


Earth. 


Liquid. 



lines. Seven distinct surfaces may be shown by 
changing the direction and angle of the section 
lines. 

Differences in the material are also indicated by 
the character of the lines, but there is no general 
agreement as to notation. Fig. 153 illustrates 



eighteen types of sectioning, together with the 
names of the materials which are indicated by 



C^ 







'^y/, 
4 
'/-' 



^?- 









w^y 



>> 






O 




::..ir 




Fig. 153. 



GENEEAL INSTRUCTIONS 



69 



them. Whenever figures or notes occur in a sec- 
tion, the section lines should not be drawn across 
them. 

Instructions for Inking. Never begin the inking 
of a drawing until the pencilling is completed. 
Always ink the circles and circular arcs first, be- 
ginning with the small arcs. If the drawing is to 
have shade lines, shade each arc as drawn. To 
omit the shading of circles until they have first 
been inked in fine lines will necessitate almost 
double the time otherwise required. It should be 
observed that the width of both fine and heavy 
lines is determined by the shading of the first 
circular arc. Next, ink all the full and dotted 
straight lines. Begin on the upper side of the 
sheet and ink all the fine horizontal lines, omitting 
those lines which are to be shaded. Next, ink the 
vertical lines, beginning with those at the left. Do 
not dwell too long at the end of a line, especially 
if it be a, heavy one, as the pressure of the ink 
in the pen will tend to widen the line. If a series 
of lines radiate from a point, allow sufficient time 
for the drying of each line, otherwise a blot may 
be made. Ink all lines at other angles and 
those curved lines requiring the use of curves. 
The same order is to be followed in the inking 
of the shade lines, evenness being secured 
by ruling them at one time. Draw the section 
lines and ink the center and the dimension lines, 
and put on the figures and notes. 



Tracings. When it is desired to reproduce a 
drawing, tracing cloth is placed above the orig- 
inal and the lines of the drawing traced on the 
new surface, as though one were inking a pencilled 
drawing. Tracing cloth is usually furnished with 
one surface glazed and the other dull. Either side 
may be used, but it is difficult to erase from the 
dull side. Pencilling must be done on the dull sur- 
face. Tracing cloth is used frequently in place of 
paper for original drawings, the pencilled draw- 
ing being traced on the cloth in ink. 

As the cloth absorbs moisture quite rapidly, it 
shrinks and swells under varying atmospheric con- 
ditions. Because of this, large drawings which 
require considerable time to complete, should be 
inked in sections, as ihe cloth will require frequent 
adjustment in order that its surface may be smooth 
and in contact with the paper drawing. Only the 
best quality of cloth should be used, as the cheaper 
kinds are improperly sized and absorb ink, caus- 
ing blots. If the ink fails to run freely on the 
cloth, dust on the surface a little finely powdered 
pumice stone or chalk, rubbing it lightly across 
the surface with a piece of chamois skin or cloth. 

Inked lines may be removed from the cloth by 
means of a sharp knife and an ink eraser, or by 
dusting a little finely powdered pumice stone on 
the lines to be erased, and briskly rubbing them 
with the end of the finger or a piece of rubber. 
As the pumice becomes discolored replace it with 



70 



MECHANICAL DEAWING 



GEOMETRIC LETTERS. 



ABCDEFDHI 

JKLMNDPQR 

STUVWXYZ 

B b c d e f g li 1 

jkliiLnapq.r 

ktuvwxyz 

1234567830 



A 
J 
S 
a 



GOTHIC LETTERS. 



B C D E F G H I 
KLMNOPQR 
T U V W X Y Z 

b c d e f g h i 
jklmnopqf 
s t u V w X y z 

12345 9 7890 

rig. 154. 



fresh powder. Pencilled lines may be removed by 
the ordinary pencil eraser, or by means of a cloth 
moistened with benzine. 

Lettering'. The subject of lettering is of such 
importance to the draftsman that he should adopt 
some clear type for general use, and acquire pro- 
ficiency in the free-hand rendering of it. While at 
times it may be necessary to make use of instru- 
ments and mechanical aids for the construction of 
letters and figures, usually they may be written 
free-hand. 

The accompanying alphabets shown in Fig. 154, 
will be acceptable in the regular practice of draft- 
ing. Their study will afford an excellent exercise, 
as well as skill in the figuring and lettering of 
drawings. Both types should be written with the 
aid of instruments. The first is known as the 
geometric, and the second as the Gothic. Large 
and small capitals may be used in the place of 
capitals and lower case, as illustrated. The small 
capitals and lower case may be made about two- 
thirds the height of the initial letters. 

Eound writing is also shown in Fig. 155, 
pens for executing this style of writing and letter- 
ing may be purchased at any drawing material 
dealer's store. 

Instructions for Pencilling. Pencilling is a pre- 
requisite to inking. Of first importance is the 
sharpening of a pencil, and as it wears away rap- 
idly it must be sharpened frequently. 



GENERAL INSTEUCTIONS 



71 



The pencil should be held vertical, or nearly 
so, the arm free from the body, and the flat edge 
of the chisel point lightly touching the straight- 
edge. Do not attempt to draw with the pencil 
point in the angle made by the paper and the edge 




^ 01OV/i/^>\AAA?^^ ^ 



Fig. 155. 

of the blade or straight-edge. Draw from left to 
right or from bottom to top of board. In general, 
lines are drawn from the body, the draftsman fac- 
ing the board when drawing horizontal lines, and 
having his right side to the board when drawing 



long vertical lines. For the drawing of other lines 
the position of the draftsman should be such as to 
enable him to draw at ease, having a free-arm 
movement, even though it be necessary to draw 
from the opposite side of the board. 

Economize in the drawing of lines by omitting 
such portions as may be unnecessary. Lines, 
which should be dotted when inking, may fre- 
quently be pencilled in full. 

Dimensioning. To dimension a drawing is to 
place upon it all measurements of the object rep- 
resented, so that the workman may construct the 
object from the given measurements. 

In order to be easily read the dimensions should 
be so placed as not to crowd or interfere with each 
other or with the lines of the drawing, and the 
figures must be neatly made, poor figures will spoil 
the appearance of the best drawing. 

Dimensions should be placed upon dimension 
lines which, by means of arrow-heads at each end, 
indicate the position of the dimension. These 
lines should not be continuous, a space being left 
to receive the figures, which should be symmet- 
rically placed upon the line. The line may be 
omitted when the space between the arrow-heads 
is short, and when there is not room for both 
arrow-heads and dimension, the arrow-heads may 
be turned in the direction of the measurement and 
placed outside the line. 

Wben there is not room even for the dimension, 



72 



MECHANICAL DRAWING 



arrow-heads may be used either outside or inside 
the dimension lines, and the dimension placed 
where there is room for it. 

Arrow-heads and figures should be drawn free- 
hand, when the drawing is inked they should be 
drawn with a common writing pen and with black 
ink, while the dimension lines should be red. The 
line separating the figures indicating the fraction 
must be parallel to the dimension line. 

Vertical dimensions should always be placed so 
as to read right-handed. 

The dimensions may be placed upon the draw- 
ing when there is room, but when the space is 
small it is better to carry the dimension outside 
the drawing by means of dot and dash lines. 

The space between the different views is often 
the best position for many of the dimensions. 

When an object is divided into different parts 
and the lengths are given in detail, an over-all 
dimension should be given. 

Dimensions should not be placed upon center 
lines. 

Distances between centers of alJ parts, such as 
rods, bolts, or any evenly spaced £axts, should be 
given, and when the parts are arranged in a circle, 
the diameter of the circle passing through their 
centers should be given. 

The diameter of a circle and the radius of an arc 
should be given. The center of an arc, when not 
otherwise shown, should be marked by a small 



circle placed about it, and used instead of an 
arrow-head. The dimension line should begin at 
this small circle. 

Dimensions should be clearly given in some one 
view, and not repeated in other views, they should 
seldom be placed between a full and a dotted line, 
or between dotted lines when they can be placed 
in a view where the part is represented by full 
lines. 

Simple objects, circular in section, are often 
shown by only one view. 

When several pieces are alike, only one is 
drawn, and the number to be made is expressed 
by lettering. 

When parts are to be fitted together, it is cus- 
tomary to write whether the fit is to be tight or 
loose. 

Working Drav/ings. A good working drawing 
should be prepared in the following manner: 
It must first be carefully outlined in pencil and 
then inked in. After this all parts cut by planes of 
section should be cross hatched, the cross hatch- 
ing used indicating the materials of which the 
parts are made. The center lines are now inked 
in with red ink. The red ink may be prepared by 
rubbing down a cake of crimson lake. Next come 
the distance or dimension lines, which should be 
put in with red ink. The arrow-heads at the ends 
of the dimension lines are now put in with black 
ink, and so are the figures for the dimensions. 



GENERAL INSTRUCTIONS 



73 



The arrow-heads and the figures should be made 
with a common writing pen. The dimensions 
should be put on neatly. Many a good drawing 
has its appearance spoiled through being slovenly 
dimensioned. 

Here may be pointed out the importance of put- 
ting the dimensions on a working drawing. If 
the drawing is not dimensioned, the workman 
must get his sizes from the drawing by applying 
his rule or a suitable scale. Now this operation 
takes time, and is very liable to result in error. 
Time is therefore saved, and the chance of error 
reduced, by marking the sizes in figures. 

Tinting". A drawing is colored or tinted for 
the purpose of making clear the divisions, as in 
map drawing, to indicate the character of the sur- 
face, whether plane or curved, and possibly its 
relation to other surfaces by the casting of shad- 
ows, or to designate the materials used. 

The paper should be of proper quality, such as 
Whatman's cold pressed, and must be stretched 
by wetting the surface and glueing it to the board 
in the following manner: Having laid the paper 
on a flat surface, fold over about one-half inch of 
each edge. Thoroughly wet all the paper ex- 
cept folded edges, using a soft sponge for this 
purpose, but do not rub the surface of the paper. 
Next apply mucilage, strong paste, or a light glue, 
to the underside of the folded portion and press 
this to the board with a slight outward pressure 



so as to bring the surface of the paper close to 
the board. As the glue should set before the 
paper begins to dry and shrink, it is necessary to 
have the paper very wet, but no puddles must be 
allowed to remain on the surface after the edge is 
glued. The paper must be allowed to dry grad- 
ually in a perfectly horizontal position, as other- 
wise the water would tend to moisten the lower 
edge and prevent the drying of the glue. If the 
paper should dry too rapidly, not allowing suffi- 
cient time for the glue to set, the surface may be 
moistened again. 

The color employed in making the wash may be 
a water color or ground India ink, but none of the 
prepared liquid inks are suitable for the purpose. 
The color should be very light, and when the de- 
sired shade is to be dark it should be obtained by 
applying several washes, allowing sufficient time 
for each to dry. The color must be free from sedi- 
ment, but since some deposit is liable to take place, 
the brush should be dipped in the clear portion 
only, and not allowed to touch the bottom of the 
saucer. 

The brush should be of good size, depending 
somewhat on the surface to be covered, and of such 
quality that when filled with the color or water, 
it will have a good point. 

Two classes of tinting are employed, the flat 
tint of uniform shade, and the graded tint for the 
representation of inclined or curved surfaces. 



74 



MECHANICAL DRAWING 



Remove all pencilled lines which are not to be 
a part of the finished drawing, and do all the 
necessary cleaning of the surface, using the great- 
est care not to roughen the paper. Inking should 
be done after the tinting, but if for any reason it 
is necessary to ink the drawing first, a water- 
proof ink must be used. 

In putting on the color, slightly incline the 
board to permit of the downward flow of the 
liquid, and, beginning at the upper portion of 
the drawing, pass lightly from left to right, using 
care just to touch the outline with the color, but 
not to overrun, and making a somewhat narrow 
horizontal band of color. Advance the color by 
successive bands, the brush just touching the 
lower edge of the pool of water made by the pre- 
ceding wash. This lower edge should never be al- 
lowed to dry, as it would cause a streak to be made 
in the tinted surface. Having reached the lower 
edge, use less water in the brush so as to enable 
a better contact to be made with the outline. 
Finally dry the brush by squeezing it or touching 
it to a piece of blotting paper. It may then be 
used to absorb the small pool of color at the bot- 
tom edge or comer. 

Avoid touching the tinted surface until it is dry, 
at which time any corrections that are necessary 
may be made by stippling. This consists in using 
a comparatively dry brush and cross hatching 
the surface to be corrected, 



If the surface to be covered is large, it is desir- 
able to apply a wash of clear water before apply- 
ing the color. This dampened surface will pre- 
vent the quick drying of the color and insure a 
more even tint. When necessary to remove the 
tint from a surface, use a sponge with plenty of 
clean water, and by repeated wettings absorb the 
color, but do not rub the surface of the paper. 

A graded tint may be applied by several meth- 
ods, the simplest being to divide the surface into 
narrow bands and apply successive washes, each 
covering an additional band. If the tint is suffi- 
ciently light, and the bands narrow, the division 
line between the bands will not be very noticeable, 
but this may be lessened by the softening of the 
edges with a comparatively dry brush and clean 
water. 

Another method, which requires considerable 
dexterity, is to put on a narrow band of the dark- 
est tint that may be required, and, instead of re- 
moving the surplus water from the edge, touch 
the brush to some clean water and with this 
lighter tint continue the wash over the second 
band. Continue in this manner until the entire 
surface is covered. 

Making Blue Prints. To make good blue prints, 
being guided only by the appearance of the ex- 
posed edge of sensitized paper, requires consid- 
erable experience. Very often, especially on a 
(^loudy day, the edge looks just about right, but 



GENERAL INSTRUCTIONS 



75 



•wlien taken out of the frame and given a rinsing, 
it is only to find that the print looks pale because 
it should have been allowed to remain exposed for 
a longer period. 

Take a small test-piece of the same paper, about 
4 inches square, and a piece of tracing cloth with 
several lines on its surface and lay these small 
pieces out at the same time the real print is being 
exposed, and cover these samples with a piece of 
glass about 4 inches square. As a general rule, 
a place can be found on top of the frame for the 
testing-piece, and by having a small dish of water 
at hand for testing the print by tearing off a 
small bit and washing same to note its appearance, 
the novice can get just as good results as the ex- 
perienced hand and without much danger of fail- 
ure. 

When several prints are to be made the second 
one may be placed into the frame while the first 
one is soaking, when the print is properly soaked, 
say about ten minutes, lift it slowly out of the 
water by grasping two of its opposite corners; 
immerse again and pull out as before. This is 
to be continued until the paper does not change 
to a deeper blue color. Hang the paper on the 
rack by two of its comers to dry. In case any 
spots appear it is an indication that the prints 
were not properly washed. 

When corrections or additions are to be made 
to a blue print a special chemical preparation must 



be used to make white lines. A solution of bicar- 
bonate of soda and water is generally used for this 
purpose. When white lines or figures are to be 
obliterated a blue pencil may be used to cover 
same. 

The solution used for ordinary blue printing is 
made according to the following receipt : 

One ounce of red prussiate of potash dissolved 
in 5 ounces of water. 

One ounce of citrate of iron and ammonia dis- 
solved in 5 ounces of water. 

Keep the solution separate in dark colored bot- 
tles in a dark place not exposed to the light. To 
prepare the paper, mix equal portions of the two 
solutions and be careful that the mixtures are not 
longer exposed to the light than is necessary to 
see by. It is, therefore, a necessity to perform this 
work in a dark room, provided with a trough of 
some kind to hold water, this should be larger 
than the blue print and from six to eight inches 
deep, a flat board should be provided to cover this 
trough, there should also be an arrangement like 
a towel rack to hang the paper on while drying. 
The sheets should be cut in such a manner as to 
be a little larger than the tracing, in order to leave 
an edge around it when the tracing is placed upon 
it. From ten to twelve sheets are placed upon a 
flat board, care must be taken to spread them flat 
one above another, so that the edges are all even. 
The sheets should be secured to the board by a 



76 



MECHANICAL DEAWING 



small nail throiigh the 'two upper comers, strong 
enough to hold the weight of the sheets when the 
board is placed vertically. 

Place the board on the edges of the trough with 
one edge against the wall and the board somewhat 
inclined, only as much light as is absolutely re- 
quired should be obtained from a lamp or gas jet, 
turned down very low. The solution referred to 
above should be applied evenly with a wide brush 



or a fine sponge over the top sheet of paper. 
When the top sheet is finished remove it from the 
board by pulling at the bottom of same and tear- 
ing it from the nail which holds it, place the sheet 
in a drawer where it can lie flat and where it 
cannot be reached by the light. 

Treat the remaining sheets in the same way as 
the first one. 



MACHINE DRAWING 



The draftsman should not as a rule be content 
with simply reproducing the views shown in the 
different examples given, to the dimensions 
marked on them, but should lay out other views 
and cross sections. The great importance of the 
value of being able to make intelligible free hand 
sketches of machine details cannot be overesti- 
mated, the draftsman should practice this art, not 
only from the illustrations given herewith, but 
from actual machine details. Fully dimensioned 
free hand sketches of actual machines or their de- 
tails, form excellent examples for drawing prac- 
tice. All such sketches should be made in a book 
kept for the purpose, always putting in the dimen- 
sions where possible. The description of the vari- 
ous applications of the mechanical powers herein- 
before given is more for reference than for the 
purpose of teaching these principles. As machine 
drawing is simply the application of the principles 
of geometry to the representation of machines, if 
the draftsman or student is not already familiar 
with the study of geometry, he should make him- 
self acquainted with the problems given in this 
work, before going further. 



U. S. Standard Hexagonal Bolt-head and Nut. 

Two types of head and nut are illustrated, the 
rounded or spherical, and the chamfered or coni- 
cal, as shown in Fig. 156. Three dimensions are 
fixed by this standard: First, the distance across 
the flats or short diameter, commonly indicated by 
H, and equal to one and one-half times the diam- 
eter of the bolt plus one-eighth of an inch, second, 
the thickness of the head, which is equal to one- 
half its short diameter, third, the thickness of the 
nut, which is equal to the diameter of the bolt. 

Example 1: Hexagonal head bolt and nut. 
Draw the views of the bolts and nuts as shown in 
Fig. 156, for bolts 4 inches long under head and 
1 inch diameter. Scale — Full size. 

Cast iron flange coupling. In the kind of coup- 
ling shown in Fig. 157 a cast iron center or boss 
provided with a flange is secured to the end of 
each shaft by a sunk key driven from the face 
of the flange. These flanges are then connected by 
bolts and nuts. 

To ensure the shafts being in line the end of one 
projects into the flange of the other. 

In order that the face of each flange may be 



77 



78 



MECHANICAL DRAWING 



exactly perpendicular to the axis of the shaft they 
should be faced in the lathe, after being keyed on 
to the shaft. 





Fig. 156 — Hexagonal-head Bolt and Nut. 

If the coupling is in an exposed position, where 
the nuts and bolt-heads would be liable to catch 
the clothes of workmen or an idle driving band 
which might come in the way, the flanges should 



be made thicker, and be provided with recesses 
for the nuts and bolt-heads. 



Dimensions of Cast-iron Flange Couplings. 1 


1 

Diameter Diameter 


Thick- 
ness of 
flange 
T 


Diameter 


Depth 


Num- 


Diam- 


Diameter 

of bolt 

circle 

C 


of shaft 
D 


of flange 


of boss 
B 


at boss 
L 


ber of 
bolts 


of bolts 
d 


IX 


7K 


% 


3% 


2% 


3 


% 


5% 


2 


8% 


ItV 


4% 


3t^ 


4 


% 


6% 


2X 


10% 


IX 


Stt 


3% 


4 


% 


8% 


3 


12% 


ItV 


6% 


4A 


4- 


1 


9% 


8X 


13% 


1% 


7% 


4% 


4 


1 


lOA 


4 


14 


1% 


8 


SyB" 


6 


1 


11% 


4X 


15% 


2 


8% 


6 


6 


1% 


12% 


5 


17% 


2% 


9il 


6% 


6 


1% 


13il 


5X 


18X 


2t7 


10% 


7% 


6 


1% 


14% 


6 


19% 


2X 


11% 


7% 


6 


1% 


16 



The projection of the shaft p varies from i/4 



inch in the small shafts to % 



inch in the large 



ones. 

Example 2. Cast-iron Flange Coupling. Draw 
the views shown in Fig. 157 of a cast-iron flange 
coupling, for a shaft 5 inches in diameter, to the 
dimensions given in the above table. Scale — 3 
inches to 1 foot. 

Proportions of Rivet Heads. The diameter of 
the snap head is about 1.7 times the diameter of 
the rivet, and its height about .6 of the diameter of 
the rivet. The conical head has a diameter twice 
and a height three quarters of the rivet diameter. 
The greatest diameter of the oval head is about 



MACHINE DRAWING 



79 



1.6, and its height .7 of the rivet diameter. The 
greatest diameter of the countersmik head may be 




Fig. 157 — Cast-iron Flange Coupling. 

one and a half, and its depth a half of the diam- 
eter of the rivet. 



Table Showing the Proportions of Single Eiveted 
Lap Joints for Various Thicknesses of Plates. 


Thickness of plates. 


5 


/8 


tV 


X 


9 


% 


1 1 


Diam. of rivets 
Pitch of rivets 


lA 


IH 


2 


2tV 


1 

2i 


ItV 
2X 


li 

21 



Distance from center of rivets to edge of plate = IM times diameter of rivets. 

Example 3. Single-riveted butt Joints. In Fig. 
158 are shown two forms of single riveted butt 



joints. One of the sectional views shows a butt 
joint with one splice plate, the other sectional 
view shows a joint with two splice plates. The 
plan view shows both arrangements. Draw all 
these views full size. 




Fig. 158 — Single Riveted Butt Joints. 

Example 4. Comer ofWrought-iron Tank. This 
exercise is to illustrate the connection of plates 
which are at right angles to one another by means 
of angle irons. Fig. 159 is a plan and elevation 



80 



MECHANICAL DRAWING 



of the comer of a wroiiglit-iron tank. The sides 
of the tank are riveted to a vertical angle iron, the 
cross section of which is clearly shown in the 
plan. Another angle iron of the same dimensions 
is used in the same way to connect the sides with 
the bottom. The sides do not come quite up to the 
comer of the vertical angle iron, excepting at the 
bottom where the horizontal angle iron comes in. 




Fig. 159 — Corner of Wrought-iron Tank. 

At this point the vertical plates meet one another, 
and the edge formed is rounded over to fit the in- 
terior of the bend of the horizontal angle iron so 
as to make the joint tight. Draw this example 
half size. 

The dimensions are as follows: angle irons 2% 
inches X 2^2 inches X% inch, plates % inch thick, 
rivets % inch diameter and 2 inches pitch. 

Example 5. Gusset Stay. In order that the flat 



ends of a steam boiler may not be bulged out by 
the pressure of the steam they are strengthened by 
means of stays. One form of boiler stay, called 
a gusset stay^ is shown in Fig. 160. This stay con- 
sists of a strip of wrought-iron plate which passes 




'A 
if 



1^ 
® 



4^- 



@ 

© 
© 




Fig. 160— Gusset Stays. 

in a diagonal direction from the flat end of the 
boiler to the cylindrical shell. One end of this 
plate is placed between and riveted to two angle 
irons which are riveted to the shell of the boiler. 



MACHINE DRAWING 



81 



A similar arrangement connects tlie other end of 
the stay plate to the flat end of the boiler. In this 
example the stay or gusset plate is % of an inch 
thick, the angle irons are 4 inches broad and 1/2 
inch thick. The rivets are 1 inch in diameter. 
The same figure also illustrates the most common 
method of connecting the ends of a boiler to the 
shell. The end plates are flanged or bent over at 




Fig. 161 — Flanged Joint for Cast-iron Plates. 

right angles and riveted to the shell as shown. 
The radius of the inside curve at the angle of the 
flange is l^A inches. Draw this example to a scale 
of 3 inches to 1 foot. 

Example 6. Flanged Joint for Cast-iron Plates. 
Draw the views shown in Fig. 161. Draw also a 



plan. The bolts and nuts must be shown in each 
view. The holes for the bolts are square, and the 
bolts have square necks. Draw this example half 
full size. 

Pillow Block. One form of pillow block is 
shown in Pig. 16'2. A is the block proper, B the 
sole-plate through which pass the holding down 
bolts. C is the cap. Between the block and the 
cap, is the brass bushing which is in halves. 

In the block illustrated the journal is lubricated 
by a needle lubricator, this consists of an inverted 
glass bottle fitted with a wood stopper, through a 
hole in which passes a piece of wire, which has one 
end in the oil within the bottle, and the other rest- 
ing on the journal of the shaft. The wire or needle 
does not fill the hole in the stopper, but if the 
needle is kept from vibrating the oil does not 
escape owing to capillary attraction. When, how- 
ever, the shaft rotates, the needle begins to vi- 
brate, and the oil runs down slowly on to the jour- 
nal, oil is therefore only used when the shaft is 
running. 

Example 7. Pillow Block for a Four-inch Shaft. 

Draw the views shown of this block in Fig. 162. 
Scale 6 inches to 1 foot. 

Proportions of Pillow Blocks. The following 
rules may be used for proportioning pillow blocks 
for shafts up to 8 inches diameter. It should be 
remembered that the proportions used by different 



8^ 



MECHANICAL DRAWING 




bO 

.a 

pi 






M 



O 



to 



bn 



makers vary considerably, but the following rules 
represent average practice: 

Diameter of journal . . . =d 

Length of journal . . . . =1 

Height to centre .... =1.05<?+.5 

Length of base .... =S.6d+5 

Width of base . . . . =.8l 

" block . . . . =.7l 

Thickness of base . . . . = .Zd+ .3 

cap .... =.Bd+.4: 

Diameter of bolts .... =.25d+.25 

Distance between centres of cap bolts =1.6^+1.5 

base bolts =2.7(^+4.2 

Thickness of step at bottom. . . z=t= 0dd+.15 

sides . . =/ii 

The length of the journal varies very much in 
different cases, and depends upon the speed of the 
shaft, the load which it carries, the workmanship 
of the journal and bearing, and the method of 
lubrication. For ordinary shafting one rule is to 
make l=d+l. Some makers use the rule l=1.5d, 
others make l=2d. 

Example 8. Sole Plate for a Pillow Block. 
Draw the views for a sole plate for a pillow block 
as shown in Fig. 163. Draw also an end elevation. 
Scale — ^Half size. 

Example 9. Bracket for Pillow Block. Draw 
the side and end elevations shown in Fig. 164, 
and from the side elevation project a plan. Scale 
— ^Half size. 



MACHINE DRAWING 



83 



Example 10. "Wall Bracket and Bearing. Draw 
the side and end elevation partly in section as 
shown in Fig. 165, and project a complete plan 
below. Scale — Half size. 

Pulleys. Let two pulleys A and B be connected 
by a belt, and let their diameters be Dj and D2; 
and let their speeds, in revolutions per minute, be 
Nj and Ng respectively. If there is no slipping, 




^ 







m^ 



^ 



'i 



-AW 
to 



34^ 



I'e^"- 




Fig. 163— Sole Plate for a Pillow Block. 

the speeds of the rims of the pulleys will be the 
same as that of the belt, and will therefore be 
equal. Now the speed of the rim of A is evi- 
dently=DiX 3.1416 XNj, while the speed of the 
rim of B 18=^2 X 3.1416x^2. Hence DjX 3.1416 
X^i=D2XS I4I6XN2, and therefore 



Pulleys for Flat Belts. In cross section the rim 
of a pulley for carrying a flat belt is generally 
curved as shown in Fig. 166, but very often the 
cross section is straight. The curved cross section 
of the rim tends to keep the belt from coming off 
as long as the pulley is rotating. Sometimes the 



6^'— --N 




Fig. 164r— Bracket for a Pillow Block. 

rim of the pulley is provided with flanges which 
keep the belt from falling off. 

Pulleys are generally made entirely of cast iron, 
but a great many pulleys are now made in which 
the center or hub only is of cast iron, the arms 
being of wrought iron cast into the hub, while the 
rim is of sheet iron. 

The arms of pulleys when made of wrought iron 
are invariably straight, but when made of cast 



84 



MECHANICAL DRAWING 



iron they axe very often curved. In Fig. 166 whicli 
shows an arrangement of two cast-iron pulleys, the 
arms are straight. Through unequal cooling, and 
therefore unequal contraction of a cast-iron pulley 
in the mould, the arms are generally in a state of 



I ->ir*^-»1 



4; 




-2^2 



.yy : — -->K- — -:-- 5^ T--r,rH 




Fig. 165 — Wall-bracket and Bearing. 



yield, and thus cause a diminution of the stress 
due to unequal contraction. 

The cross section of the arms of cast-iron pul- 
leys is generally elliptical. 

Example 11. Tight and Loose Pulleys. Fig. 
166 shows an arrangement of tight and loose pul- 




Fig. 166— Tight and Loose Pulleys. 



tension or compression, and if the arms are 
straight they are very unyielding, so that the re- 
sult of this initial stress is often the breaking of 
an arm, or of the rim where it joins an arm. With 
the curved arm, howeverj its shape permits it to 



leys. A is the last pulley, secured to the shaft C 
by a sunk key, B is the loose pulley, which turns 
freely upon the shaft. The loose pulley is pre- 
vented from coming off by a collar D, which is 
secured to the shaft by a tapered pin as shown. 



MACHINE DRAWING 



85 



The nave ot boss of the loose pulley is here fitted 
with a brass bushing, which may be renewed when 
it becomes too much worn. Draw the elevations 
shown, completing the left-hand one. Scale 6 
inches to 1 foot. 

By the above arrangement of pulleys a machine 
may be stopped or set in motion at pleasure. 
"When the driving belt is on the loose pulley the 
machine is at rest, and when it is on the tight pul- 
ley the machine is in motion. The driving belt 
is shifted from the one pulley to the other by 
pressing on that side of the belt which is advanc- 
ing towards the pulleys. 

Gear Wheels. Let two smooth rollers be placed 
in contact with their axes parallel, and let one of 
them rotate about its axis, then if there is no slip- 
ping the other roller will rotate in the opposite di- 
rection with the same surface velocity, and if Dj, 
Dz be the diameters of the rollers, and N^, Ng their 
speeds in revolutions per minute, it follows as 
in belt gearing that 

If there be considerable resistance to the motion 
of the follower slipping may take place, and it 
may stop. To prevent this the rollers may be pro- 
vided with teeth, then they become spur wheels, 
and if the teeth be so shaped that the ratio of the 
speeds of the toothed rollers at any instant is 



the same as that of the smooth rollers, the sur- 
faces of the latter are called the pitch surfaces of 
the former. 

Pitch Circle. A section of the pitch surface of 
a toothed wheel by a 'plane perpendicular to its 
axis is a circle, and is called a pitch circle. We 
may also say that the pitch circle is the edge of 
the pitch surface. The pitch circle is generally 
traced on the side of a toothed wheel, and is rather 
nearer the points of the teeth than the roots. 

Pitch of Teeth. The distance from the center of 
one tooth to the center of the next, or from the 
front of one to the front of the next, measured at 
the pitch circle, is called the pitch of the teeth. 
If D be the diameter of the pitch circle of a wheel, 
n the number of teeth, and p the pitch of the teeth, 
then Dx3.1416=nXp. 

By the diameter of a wheel is meant the diam- 
eter of its pitch circle. 

Form and Proportions of Teeth. The ordinary 
form of wheel teeth is shown in Fig. 167. The 
curves of the teeth should be cycloidal curves, al- 
though they are generally drawn in as arcs of cir- 
cles. It does not fall within the scope of this 
work to discuss the correct forms of gear teeth. 

Example 12. Spur Gear. Fig. 167 shows the 
elevation and sectional plan of a portion of a 
cast-iron spur gear. The diameter of the pitch 
circle is 23y8 inches, and the pitch of the teeth 
is 1% inches, so that there will be 50 teeth in the 



86 



MECHANICAL DRAWING 



gear. Tlie gear has six arms. Draw a complete 
elevation of the gear and a half sectional plan, 
also a half-plan without any section. Draw also 
a cross section of one arm. Scale 3 inches to 1 
foot. 




rig. 167 — Portion of a Cast-iron Spur Gear. 

Mortise Gears. When two gears meshing to- 
gether run at a high speed the teeth of one are 
made of wood. Tliese teeth, or cogs, as they are 
generally called, have tenons formed on them, 
which fit into mortises in the rim of the gear. This 



gear with the wooden teeth is called a mortise 
gear. 

Coupling Rods. A rod used to transmit the mo- 
tion of one crank to another is called a coupling 
rod. A familiar example of the use of coupling 
rods will be found in the locomotive. Coupling 
rods are made of wrought iron or steel, and are 
generally of rectangular section. The ends are 
now generally made solid and lined with solid 
brass bushes, without any adjustment for wear. 
This form of coupling rod end is found to answer 
very well in locomotive practice where the work- 
manship and arrangements for lubrication are ex- 
cellent. When the brass bush becomes worn it is 
replaced by a new one. 

Fig. 168 shows an example of a locomotive coup- 
ling rod end for an outside cylinder engine. In 
this case it is desirable to have the crank-pin 
bearings for the coupling rods as short as possible, 
for a connecting rod and coupling rod in this kind 
of engine work side by side on the same crank- 
pin, which, being overhung, should be as short as 
convenient for the sake of strength. The requisite 
bearing surface is obtained by having a pin of 
large diameter. The brass bush is prevented from 
rotating by means of the square key shown. The 
oil-box is cut out of the solid, and has a wrought- 
iron cover slightly dovetailed at the edges. This 
cover fits into a check round the top inner edge of 
the box, which is originally parallel, but is made 



MACHINE DRAWING 



87 



to close on the dovetailed edges of the cover by 
riveting. A hole in the center of this cover, which 
gives access to the oil-box, is fitted with a screwed- 
brass plug. The brass plug has a screwed hole in 




1^ gii ,1 

Fig. 168 — Locomotive Coupling-rod End. 

the center, through which oil may be introduced 
to the box. Dust is kept out of the oil-box by 
screwing into the hole in the brass plug a common 



cork. The oil is carried slowly but regularly 
from the oil-box over to the bearing by a piece 
of cotton wick. 




rig. 169— Gland and Stufang-bos. 



88 



MECHANICAL DEAWING 



nn 

rin K7,^f - H fjln 



rn 




Fig. 170— Water or Steam Cock 



■ !■ 




1 

-J 


^2? 
1 






1 
1 
1 




± 



Example 13. Coupling Rod End. Draw first the 
side elevation and plan, each partly in section as 
shown in Fig. 168, Then instead of the view to the 
left, which is an end elevation partly in section, 
draw a complete end elevation looking to the right, 
and also a complete vertical cross section through 
the center of the bearing. Scale 6 inches to 1 foot. 

Stuffing-boxes. In Fig. 169 is shown a gland 
and stnfifing-box for the piston rod of a vertical 
engine. A B is the piston rod, C D a portion of the 
cylinder cover, and E F the stuffing-box. Fitting 
into the bottom of the stuffing-box is a brass bush 
H. The space K around the rod A B is filled with 
packing, of which there is a variety of kinds, the 
simplest being greased hempen rope. The pack- 
ing is compressed by screwing down the cast-iron 
gland L M, which is lined with a brass bush N. 
In this case the gland is screwed down by means of 
three stud-bolts P, which are screwed into a flange 
cast on the stuffing-box. Surrounding the rod on 
the top of the gland there is a recess R for holding 
the lubricant. 

The object of the gland and stuffing-box is to 
allow the piston rod to move backwards and for- 
wards freely without any leakage of steam. 

Example 14. Gland and Stuffing-box for a Ver- 
tical Rod. Draw the views shown in Fig. 169 to 
the dimensions given. Scale 6 inches to 1 foot. 

Water or Steam Cock. Fig. 170 shows a cock 
of considerable size, which may be used for water 



MACHINE DRAWING 



89 







o 






bo 

•l-l 



or steam under high pressure. The plug in this 
example is hollow, and is prevented from coming 
out by a cover which is secured to the casing by 




«- — 


f 

_ L - 

1 
J 

1 
1 

1 
1 

1 


-A" 





--> 











Fig. 172— Details of Tailstocls., 



90 



MECHANICAL DRAWING 



four stud bolts. An annular ridge of rectangular 
section projecting from the under side of the 
covei', and fitting into a corresponding recess on 
the top of the casing, serves to ensure that the cov- 
er and plug are concentric, and prevents leakage. 







S guar A threads Left hand. 



Ft ue I A reads 



eads h e 



er I 



nc/i. 



<^^ 



fieafAer if^y ^ 



3^ 

r iey) 







Vjao 



c^^J. 



Fig. 173— Details of Tailstock. 

Leakage at the neck of the plug is prevented by 
a gland and stuffing-box. The top end of the plug 
is made square to receive a handle for turning it. 
The size of a cock is taken from the bore of the 
pipe in which it is placed, thus Fig. 170 shows a 
2%-inch cock. 






Fig. 174— Surface Gauge. 



MACHINE DRAWING 



91 



Example 15. 214-inch Steam or Water Cock. 

First draw the views of this cock shown in Fig. 
170, then draw a half end elevation and half cross 
section through the center of the plug. Scale 6 
inches to 1 foot. 







rig. 175 — Details of Surface Gauge. 

Instead of drawing the parts of the pipe on the 
two sides of the plug in the same straight line as 
in Fig. 170, one may he shown proceeding from 
the bottom of the casing, so that the fluid will 
have to pass through the bottom of the plug and 
through one side. This is a common arrangement. 



All the parts of the valve and casing in this ex- 
ample are made of brass. 

Example 16. Tailstock for 12-inch Lathe. Two 
views of this tailstock are shown in Fig. 171. On 




Fig. 176— Details of Surface Gauge. 

one of these views a few of the principal dimen- 
sions are marked. The details, fully dimensioned, 
are shown separately in Figs. 172 and 173, 

Explain clearly how the center is moved back- 
wards and forwards, and also how the spindle con- 



92 



MECHANICAL DRAWING 



taining it is locked when it is not required to 
move. 

Draw, half-size, the views shown in Fig. 171, 
and from the left-hand view project a plan. Draw 




Fig. 177— Jack Screw. 



also the detail of the locking arrangement shown 
in Fig. 172. 
Example 17. Surface Gauge. First draw, full 



A 





"^ @ 
! L 




HaXJ sectioTutl Plan at A^,. 
Fig. 178— Reversible Ratchet-drill. 



size, all the details separately, as shown in Figs. 
175 and 176, then draw, full size, the plan and 
two elevations of the tool complete, as shown in 
Fig. 174. 



MACHINE DRAWING 



93 



F is the scriber which may be clamped at any 
part of the straight portion, between D and E. 
The scriber may also be placed at any angle to 



K— - 








Fig. 179— Details of Ratchet-drill. 

the horizontal, and the point at which it is 
clamped may be placed at any height from the 




Fig. 18Q-^§team Ingijie Governor, 



94 



MECHANICAL DRAWING 



base A within the limits of the upright K. D 
and E are carried by the clamp H, which embraces 
the upright K. By turning the milled nut J the 
scriber is fixed in position in relation to K. A 




fine vertical adjustment is obtained by rotating 
the milled nut B. After all adjustments have been 
made, K is locked in position by the set-screw C. 
Example 18. Jack Screw. From the half eleva- 



13 



H- zYz- 

Tig. 181— Details 




of Governor 



Tig. 182 — Details of Governor. 



MACHINE DRAWING 



95 



tion and half section shown in Fig. 177 make 
working drawings of the separate parts of the 
screw-jack, then draw the views shown below of 
the complete machine. Scale 6 inches to 1 foot. 



-.V-*. 







Pig. 183— Details of Governor. 



Example 19. Reversible Ratchet-drill. Draw, 

full size, the views shown in Fig, 178 of a rever- 
sible ratchet-brace. Draw also the details sep- 
arately, as shown in Fig. 179. All the dimensions 
are to be obtained from the detailed drawings. 




H^hi 



.^ holes 



^ H ^Wf kJ^ 



Fig. 184 — Details of Grovernor. 



Example 20. Steam Engine Grovemor. Draw 

the half elevation and half section of the governor 
complete, as shown in Fig. 180. Draw also a plan 



96 



MECHANICAL DRAWING 



and an elevation looking in the direction of tlie 
arrow (a). Scale 6 inches to a foot. All the di- 
mensions are tO' be taken from the illustrations 
of the details shown in Figs. 181, 182, 183 and 
184. 

The governor illustrated is used on an engine 
having a cylinder 8 inches in diameter, with a pis- 



ton stroke of 16 inches. The crank-shaft runs at 
about 110 revolutions per minute, and the gover- 
nor-spindle is driven at three times the speed of 
the crank-shaft. The governor controls the ex- 
pansion valve. This type of governor is known 
as the ''Porter" governor, from the name of the 
inventor. 



TECHNICAL DEFINITIONS 



The unit of work is the work done in lifting one 
pound through a height of one foot, or work done 
when a resistance of one pound is overcome 
through a space of one foot, and is called the foot- 
pound. 

The number of units of work performed=PxS 
where P equals the force applied, or the resistance 
overcome in pounds, and S equals the space moved 
over in feet. 

Force is any action which can be expressed sim- 
ply by weight, and which can be realized only by 
an equal amount of reaction, and is the first ele- 
ment in dynamics. All bodies in nature possess 
the incessant virtue of attracting one another by 
gravitation, which action is recognized as force. 

Velocity is speed or rate of motion, and is the 
second element in dynamics. 

Time implies a continuous perception recog- 
nized as duration, or that measured by a clock, and 
is the third element in dynamics. 

Power is the product of force and velocity, that 
is to say^ a force multiplied by the velocity with 
which it is acting is the power in operation. Power 
is the differential of work, or any action that pro- 



duces work, whether mental or physical. Power 
multiplied by the time of action is work, work 
divided by time is power. 

Work is the product obtained by multiplying 
together the three simple elements, force, velocity, 
and time. 

The energy of a body is its capacity for per- 
forming work. 

The potential energy, or the energy stored in 
a body, is the product of the effort and the dis- 
tance through which it is capable of acting. 

Kinetic energy, or accumulated work of a mov- 
ing body, is the product of the mass and half the 
square of its velocity, or the weight of the body 
multiplied by the height from which it must fall 
to gain its velocity. 

Gravity is the mutual tendency which all bodies 
of nature have to approach each other, or the ten- 
dency of any falling body to approach the center 
of the earth. 

The mass of a body is its weight divided by 32.2. 

The acceleration of motion is the rate of change 
in the velocity of a moving body, which is in- 
creased at different intervals of time. 



97 



98 



MECHANICAL DRAWING 



The unit of acceleration is tliat which imparts 
unit change of velocity to a moving body in unit 
time, or an acceleration of one foot per second in 
one second. 

The acceleration due to gravity varies at dif- 
ferent places on the earth's surface. It is reck- 
oned at 32.2 feet per second in this country, and is 
generally indicated by the letter g. 

Retarded motion. The motion of a body, in- 
stead of being accelerated, may be retarded, that 
is, its velocity may decrease at different intervals 
of time. 

Varied motion is usually understood to refer to a 
moving body, when the change varies in either ac- 



celerated or retarded motion, at different intervals 
of time. 

Inertia is that quality inherent in matter where- 
by it is absolutely passive or indifferent to a state 
of motion. 

A couple consists of two parallel forces which 
are equal, and act in opposite directions. 

The weight of a body is the pressure which the 
mutual attraction of the earth and the body causes 
that body to exert on another with which it is in 
contact^mass multiplied by 32.2. 

Linear Velocity is the rate of motion in a 
straight line, and is measured in feet per second, 
or per minute, or in miles per hour. 



MATERIALS USED IN MACHINE CONSTRUCTION 



Cast Iron. The essential constituents of cast 
iron are iron and carbon, the latter forming from 
2 to 5 per cent, of the total weight. Cast iron, how- 
ever, usually contains varying small amounts of 
silicon, sulphur, phosphorus, and manganese. 

In cast iron the carbon may exist partly in the 
free state and pai-tly in chemical combination with 
the iron. 

Chilled Castings. When grey cast iron is melted 
a portion of the free carbon combines chemically 
with the iron, this, however, separates out again if 
the iron is allowed to cool slowly, but if it is sud- 
denly cooled a greater amount of the carbon re- 
mains in chemical combination, and a whiter and 
harder iron is produced. Advantage is taken of 
this in making chilled castings. In this process 
the whole or part of the mould is lined with cast 
iron, which, being a comparatively good conductor 
of heat, chills a portion of the melted metal next to 
it, changing it into a hard white iron to a depth 
varying from % to I/2 an inch. To protect the cast- 
iron lining of the mould from the molten metal it 
is painted with loam. 

Malleable Cast Iron. This is prepared by im- 
bedding a casting in powdered red hematite, an 



99 



oxide of iron, and keeping it at a bright red heat 
for a length of time varying from several hours to 
several days according to the size of the casting. 
By this process a portion of the carbon in the cast- 
ing is removed, and the strength and toughness of 
the latter become more like the strength and 
toughness of wrought iron. 

Wrought Iron. This is nearly pure iron, and 
is made from cast iron by the puddling process, 
which consists chiefly of raising the cast iron to a 
high temperature in a reverberatory furnace in 
the presence of air, which unites with the carbon 
and passes off as gas. In other words the carbon 
is burned out. The iron is removed from the pud- 
dling furnace in soft spongy masses called blooms, 
which are subjected to a process of squeezing or 
hammering called shingling. These shingled 
blooms still contain enough heat to enable them to 
be rolled into rough puddled bars. These puddled 
bars are of very inferior quality, having less than 
half the strength of good wrought iron. The pud- 
dled bars are cut into pieces which are piled to- 
gether, reheated, and again rolled into bars, which 
are called merchant bars. This process of piling, 
reheating, and re-rolling may be repeated several 



LOFC. 



100 



MECHANICAL DRAWING 



times, depending on the quality of iron required. 
Up to a certain point the quality of the iron is im- 
proved by reheating and rolling' or hammering, but 
beyond that a repetition of the process diminishes 
the strength of the iron. 

The process of piling and rolling gives wrought 
iron a fibrous structure. When subjected to vibra- 
tions for a long time, the structure becomes crys- 
talline and the iron brittle. The crystalline struc- 
ture induced in this way may be removed by the 
process of annealing, which consists in heating the 
iron in a furnace, and then allowing it to cool 
slowly. 

Forging^ and Welding. The process of pressing 
or hammering wrought iron when at a red or white 
heat into' any desired shape is called forging. If at 
a white heat two^ pieces of wrought iron be brought 
together, their surfaces being clean, they may be 
pressed or hammered together, so as to form one 
piece. This is called welding, and is a very valua- 
ble property of wrought iron. 

Steel. This is a compound of iron with a small 
percentage of carbon, and is made either by add- 
ing carbon to wrought iron, or by removing some 
of the carbon from cast iron. 

In the cementation process, bars of wrought iron 
are iinbeddied in powdered charcoal in a fireclay 
trough, and kept at a high temperature in a fur- 
nace for several days. The iron combines with a 
portion of the carbon to form blister steel, so 



named because of the blisters which are found on 
the surface of the bars when they are removed 
from the furnace. 

The bars of blister steel are broken into pieces 
about 18 inches long, and tied together in bundles 
by strong steel wire. These bundles are raised to 
a welding heat in a furnace, and then hammered 
or rolled into^ bars of shear steel. 

To form cast steel the bars of blister steel are 
broken into pieces and melted into crucibles. 

In the Siemens-Martin process for making steel, 
cast and wrought iron are melted together on the 
hearth of a regenerative gas-furnace. 

Bessemer steel is made by pouring melted cast 
iron into a vessel called a converter, through which 
a blast of air is then urged. By this means the 
carbon is burned out, and comparatively pure iron 
remains. To this is added a certain quantity of 
spiegeleisen, which is a compound of iron, carbon, 
and manganese. 

Hardening and Tempering of Steel. Steel, if 
heated to redness and cooled suddenly, as by im- 
mersion in water, is hardened. The degree of 
hardness produced varies with the rate of cooling; 
the more rapidly the heated steel is cooled, the 
harder does it become. Hardened steel is softened 
by the process of annealing, which consists in heat- 
ing the hardened steel to redness, and then allow- 
ing it to cool slowly. Hardened steel is tempered, 
or has its degree of hardness lowered, by being 



MATERIALS USED IN MACHINE CONSTRUCTION 



101 



heated tO' a temperature considerably below that 
of a red heat, and then cooling suddenly. The 
higher the temperature the hardened steel is raised 
to, the lower does its temper become. 

Case-hardening. This is the name given to the 
process by which the surfaces of articles made of 
wrought iron are converted into steel, and con- 
sists in heating the articles in contact with sub- 
stances rich in carbon, such as bone-dust, horn 
shavings, or yellow prussiate of potash. This 
process is generally applied to the articles after 
they are completely finished by the machine tools 
or by hand. The coating of steel produced on the 
article by this process is hardened by cooling the 
article suddenly in water. 

Copper. This metal has a reddish brown color, 
and when pure is very malleable and ductile, either 
when cold or hot, so that it may be rolled or ham- 
mered into' thin plates, or drawn intO' wire. Slight 
traces of impurities cause brittleness, although 
from 2 to 4 per cent, of phosphorus increases its 
tenacity and fluidity. Copper is a good conductor 
of heat and of electricity. Copper is largely used 
for making alloys. 

Alloys. Brass contains two parts by weight of 
copper to one of zinc. Muntz metal consists of 
three parts of copper to two of zinc. Alloys con- 
sisting of copper and tin are called bronze or gun- 
metal. Bronze is harder the greater the propor- 
tion of tin which it contains, five parts of copper 



to one of tin produce a very hard bronze, and ten 
of copper to one of tin is the composition of a soft 
bronze. Phosphor bronze contains copper and tin 
with a little phosphorus, it has this advantage over 
ordinary bronze, that it may be remelted without 
deteriorating in quality. This alloy also has the 
advantage that it may be made to possess great 
strength accompanied with hardness, or less 
strength with a high degree of toughness. 

Wood. In the early days of machines wood was 
largely used in their construction, but it is now 
used to a very limited extent in that direction. 
Beech is used for the cogs of mortise gears. Yel- 
low pine is much used by pattern-makers. Box, a 
heavy, hard, yellow-colored wood, is used for the 
sheaves of pulley blocks, and sometimes for bear- 
ings in machines. Lignum-vitae is a very hard 
dark-colored wood, and remarkable for its high 
specific gravity, being 1 1-3 times the weight of the 
same volume of water. This wood is much used 
for bearings of machines which are under water. 

Shafting. Shafting is nearly always cylindrical 
and made of wrought iron or steel. Cast iron is 
rarely used for shafting. 

Axles are shafts which are subjected to bending 
without twisting. 

The parts of a shaft or axle which rest upon 
the bearings or supports are called journals, piv- 
ots, or collars. 

In journals the supporting pressure is at right 



102 



MECHANICAL DEAWiNG 



angles to the axis of the shaft, while in pivots and 
collars the pressure is parallel to that axis. 

Shafts may be solid or hollow. Hollow shafts 
are stronger than solid shafts for the same weight 
of material. Thus a hollow shaft having an exter- 
nal diameter of 10^/4 inches and an internal diame- 
ter of 7 inches would have about the same weight 
as a solid shaft of the same material 7^2 inches in 
diameter, but the former would have about double 
the strength of the latter. Hollow shafts are also 
stiffer and yield less to bending action than solid 
shafts, which in some cases, as in propeller shafts, 
is an objection. 

Twisting Moment. Let a shaft carry a lever, 
wheel, or pulley of radius E- inches, and let a force 
of P lbs. act at the outer end of the radius, and at 
right angles to it. The force P produces a twisting 
action on the shaft, which is measured by the prod- 
uct PXR. This product PxR is called the twist- 
ing moment or torque on the shaft ; and if P is in 
lbs. and R in inches, the twisting moment is PxR 
inch-pounds. The twisting moment in foot-pounds 
is got by dividing the twisting moment in inch- 
pounds by 12. 

If the shaft makes N revolutions per minute, the 
horse-power which is being transmitted is 



2xRx3.1416xPxN 2x3.1416xTxN 
- or 



12x33,000 



12x33,000 



where T is the twisting moment in inch-pounds. 

Resistance of Shafts to Torsion. The resistance 
of a shaft to torsion is directly proportional to the 
cube of its diameter. Thus if the diameter be 
doubled, the strength is increased eight (2^) times. 
Let there be two shafts of the same material, and 
having diameters Dj and D2; and let the twisting 
moments which they support when strained to the 
same extent be Ti and T2 respectively; then 
T, :% :: D/ : J)/, or TJ)/=T,J),K 

Moment of Resistance. The stress produced in 
a shaft which is subjected to twisting is a shearing 
stress. This stress is not uniform, being greatest 
at the outside of the shaft and diminishing uni- 
formly towards the center, where it is nothing. Let 
f be the greatest shearing stress on the shaft and d 
its diameter. Then the moment of resistance of 
the shaft to torsion which balances the twisting 
moment is 

3.1416 ,,. ., X m TiT> 3.1416 ,3. 
-^g d'f, so that T = PR=-^^c?3/. 

In determining the safe twisting moment for a 
mild steel shaft, f may be taken equal to 9000 lbs. 
per square inch. 



MACHINE DESIGN 



Box Couplings. The coupling illustrated in Fig. 
185 consists of a solid box made of cast-iron, bored 
out to fit the shafts, whose ends are made to butt 
together inside the box. The box may be secured 
to the shafts by means of a sunk key which ex- 
tends the whole length of the box, A better ar- 
rangement is to use two keys, both driven from the 
same end of the box, as shown in Fig. 185. With 




Fig 185— Box Shaft Coupling. 

two keys it is not so important that the keyways 
in the shafts have exactly the same depth. Two 
keys may be more tightly driven in, and are easier 
to drive bacb than a single key of the same total 
length. There must, however, be a clearance space 
between the head of the forward key, and the 
point of the hind one, so as to ensure that the 
latter is driven in with the same degree of tight- 



ness as the former. In driving the keys back also, 
they can be started separately, and therefore more 
easily, when this clearance space is allowed. Fig. 
185 shows the ends of the shafts enlarged where 
the keyways are cut, so that the latter do not 
weaken the shafts, the amount of the enlargement 



-c "., 




«r L 



103 



Fig. 186— Split Shaft Coupling. 

being such that the bottom of the keyway touches 
the outside of the shaft. 

Split Couplings. This form of coupling, which 
is shown in Fig. 186, is very easily put on or taken 
off, it has no projecting parts, the bolts being com- 
pletely covered, and, like the solid couplings, it 
may be used as a driving pulley, or a driving pul- 
ley may be placed on it. In making this coupling 



104 



MECHANICAL DRAWING 



the faces for the joint between the two halves of 
the box are first planed. The bolt-holes are then 
drilled and the two halves bolted together with 
pieces of paper between them; then the muff is 
bored out to the exact size of the shaft. When 




Fig. 187 — Cast-iron Flange Coupling. 

the paper is removed and the box put on the shaft 
and bolted up, the box gripes the shaft firmly. The 
key has no taper and should fit on the sides only. 



Dimensions of Split Couplings. 


Diam. of shaft, D 


IX 


IX 


2 


2% 


2X 


2% 


3 


3X 


4 
lOX 


Diam. of Box.Di 


4X 


5K 


5% 


6 


6% 


7Ji 


7% 


9% 


Length of Box, L 


6 


7 


8 


9 


10 


11 


12 


14 


16 


Diam. of Bolts, d 


X 


% 


% 


% 


% 


X 


Vb 


% 


Vs 


Number of Bolts 


4 


4 


4 


4 


4 


4 


4 


6 


6 



Cast-iron Flange Coupling. In this form of 
coupling, which is shown in Fig. 187, there are 
cast-iron pieces or bosses, provided with flanges 
which are keyed to the ends of the shafts to be 
connected. These flanges are fastened together by 
means of bolts and nuts as shown. Sometimes the 
shaft is enlarged where it enters the coupling, so 
as to allow for the weakening effect of the key- 
way, but more frequently it is parallel throughout, 
or very slightly reduced, as shown in Fig. 187, so 
as to form a shoulder, which prevents the shaft 
going farther into the coupling. 



Dimensions of Cast-Iron Flange Couplings. 


Diam. 

of 
Shaft. 

D. 


Diam. 


Thick- 
ness of 
Flange 
T. 


Diam. 


Depth 


Num- 


Diam. 


Diameter 


of Flange 
F. 


of Boss 
B. 


at Boss 
L. 


ber of 
Bolts. 


of Bolts 
d. 


of Bolt 

Circle 

C. 


1 


5 5-8 


3-4 


2 5-8 


2 1-16 


3 


1-2 


4 1-4 


1 1-4 


6 13-16 


13-16 


3 1-16 


2 5-16 


3 


5-8 


5 1-16 


1 1-3 


7 1-4 


7-8 


3 1-2 


2 5-8 


3 


5-8 


5 1-2 


1 3-4 


7 11-16 


1 


3 15-16 


2 7-8 


4 


5-8 


5 15-16 


2 


8 7-8 


1 1-16 


4 3-8 


3 3-16 


4 


3-4 


6 3-4 


2 1-4 


9 1-16 


1 1-8 


4 3-4 


3 7-16 


4 


3-4 


7 1-8 


2 1-3 


10 9-16 


1 1-4 


5 5-16 


3 3-4 


4 


7-8 


8 1-8 


2 8-4 


11 


1 5-16 


5 3-4 


41-16 


4 


7-8 


8 9-16 


3 


12 3-8 


1 7-16 


6 1-4 


4 5-16 


4 




9 1-2 


3 1-4 


13 5-8 


1 1-2 


6 5-8 


4 5-8 


4 




9 13-16 


3 1-2 


13 1-8 


1 5-8 


7 1-8 


4 7-8 


4 




10 5-16 


3 3-4 


13 9-16 


1 11-16 


7 9-16 


5 3-16 


4 


1 ' 


10 3-4 


4 


14 


1 3-4 


8 


5 7-16 


6 




11 1-4 


4 1-4 


14 716 


1 7-8 


8 7-16 


5 3-4 


6 




11 5-8 


4 1-2 


15 5 8 


2 


8 7-8 


6 


6 


1 1-8 


12 1-2 


4 3-4 


16 1-8 


2 1-16 


9 3-8 


6 5-16 


6 


1 1-8 


13 


5 


17 5-16 


2 1-8 


9 13 16 


6 5-8 


6 


1 1-4 


13 13-16 


5 1-4 


17 3-4 


2 1-4 


10 1-4 


6 7-8 


6 


1 1-4 


14 1-4 


5 1-3 


18 3-16 


2 5-16 


10 3-4 


7 1-4 


6 


1 1-4 


14 11-16 


5 3-4 


19 1-3 


2 7-16 


11 1-4 


7 7-16 


6 


1 3-8 


15 5-8 


6 


19 7-8 


3 1-2 


11 5-8 


7 3-4 


6 


1 3-8 


16 



MACHINE DESIGN 



105 



Gib-Heads on Keys. When the point of a key 
cannot be conveniently reached for the purpose of 
driving it out, the head should be formed as shown 
in Fig. 188. This is known as a gib-head. 




Fig. 188— Gib-head Key. 

Sliding or Feather Keys. The function of a 
sliding or feather key is to secure a piece to a 
shaft, so far as to prevent the one from rotating 
without the other, but at the same time allow of a 
relative motion in the direction of the axis of the 
shaft. This form of key has no taper, and it is 




189 — Sliding Feather-keys. 



generally secured to the piece carried by the shaft, 
in which case it is made a sliding fit in the key- 
way of the shaft. But the feather-key may be fixed 



to the shaft and made a sliding fit in the piece car- 
ried by it. 

Examples of feather-keys which slide in the key- 
way of the shaft are shown in Fig. 189. In the 
left-hand view the key has a projecting pin, A, 
formed on it, which enters a corresponding hole 
in the piece B carried by the shaft. In the other 
two views gib-heads, C, are formed on the ends of 
the key, which serve the same purpose as the pin 




Fig. 190— Fixed Feather-keys. 

A in the left-hand view, namely, to ensure that the 
key and the piece B remain or move together. 

Fig. 190 shows two examples of feather-keys 
which are fixed to the shaft. In the left-hand view 
the key is sunk into the shaft, and in the right- 
hand view the key is either a saddle-key or a key 
on a flat, and is secured to the shaft by screws as 



106 



MECHANICAL DRAWINa 



shown. In these cases the key must be long enough 
to permit of the necessary sliding motion. 



Dimensions 


OP Keys. 


D — diameter of shaft. 






B=breadth of key. 






T=thickness of sunk key. 




Ti=thickness of flat key, also= 


^thickness of sad- 


die-key. Taper of key %-inch per foot of length.l 


D. 


B. 


T. 


Ti 


D. 


B. 


T. 


Ti 


% 


tV 


y^ 


A 


4 


1% 


X 


tV 


1 


% 


y4. 


t\ 


4X 


IM 


1 6 


X 


IM 


7 


^ 


t\ 


5 


1% 


6/ 


X 


\% 


X 


5 
IT 


A 


5% 


IX 


1 1 

TT 


1 6 


1% 


9 


5 


>i 


6 


1% 


% 


X 


2 


% 


5 


>i 


7 


IX 


1 3 
TT 


1 1 

TT 


2% 


w 


% 


y 


8 


2X 


1 5 
TT 


% 


'2X 


% 


% 


5 


9 


2% 


1 


X 


2% 


il 


tV 


5 
T'B' 


10 


2X 


ItV 


il 


3 


% 


t'. 


6 
TT 


11 


2X 


lA 


ItV 


i% 


1 


y 


% 


12 


3X 


IX 


iX 












n_n_n,n_;i 




«3BzST:KBVl<ni 



The proportions of the gib-head for a key, Fig. 
188, may be h=thickness of key, l=l%h. 

Screv/ed Cast-iron Flanges for Wrought-Iron 
Pipes. In a paper read before the American So- 
ciety of Mechanical Engineers, Mr. E. F. C. Davis 
described the joint used for the steam-pipes in the 
coal mines of the Philadelphia and Eeading Coal 
and Iron Company. The pipes are of wrought-iron, 
and carry cast-iron flanges of the form shown in 



Fig. 191. The flange is screwed tightly on the pipe, 
and the end of the latter is faced off flush with the 
facing piece, which is cast on the former. One of 
the two flanges which are to be bolted together has 
cast on its face a number of lugs, which have their 
inner faces bored to fit the facing piece on the 
other flange, and thus ensure the pipes being in 
line. The joint is made steam-tight by a gum- 
joint ring which is placed between the abutting 
ends of the pipes and within the circle of the lugs, 
so that the latter keep the ring central. 

The following table gives the dimensions in 
inches adopted for this pipe joint : 



A. 
3 


B. 


N,. 


C. 


D. 


E. 


F. 


G. 


N3. 


H. 


J. 


K. 


1% 


4 


6 


Va, 


5 


u 


^ 


4 


V^ 


u 


3^ 


8'^ 


4 


634 


H 


5^ 


H 


% 


4 


Vs 


u 


Vz 


4 


9'^ 


4 


7/. 


% 


6 


% 


Vz 


4 


Vh 


% 


Vz 


5 


10!^ 


4 


8J^ 


% 


7 


Vs 


J4 


4 


'Vh 


H 


Vz 


6 


13 


6 


10 


7/8 


8 


1 


V,. 


4 


Vh 


% 


Vz 


7 


13 


6 


11 


V, 


9 


1 


Vz 


4 


% 


% 


K 


8 


14 


6 


13 


% 


10 


^% 


Vz 


6 


y% 


1 


% 


10 


16X 


8 


14 


1 


12 


IX 


Vz 


6 


y% 


1 


H 



Ni=number of bolts. N2=number of lugs. 

Jaw Coupling. For large and slow moving 
shafts, the cast-iron jaw coupling shown in Fig. 
192 is very simple and effective. In Fig. 192 one- 
half of the coupling is shown fitted with a feather- 
key, so that this half may be disconnected from the 
other or geared with it at pleasure. If the jaws 



MACHINE DESIGN 



107 



are shaped as shown in Fig. 193, the two halves of 
the coupling are more easily put into gear, but in 
this case the motion must always take place in the 
same direction. 




Pig. 191 — Cast-iron Flanges for Wrought-iron Pipes. 




hT>^-E--^T*' 



W- L • -^^ T->*- -L ->' 



Fig. 192 — Square-jaw Detachable Coupling. 

Band Friction Clutch. In the friction clutch, 
shown in Fig. 194, there is a sleeve. A, which is 
keyed to the shaft and also attached to the friction 
band, B, by means of bolts, as shown. The friction 
band, B, encircles the boss, C, of the wheel or pul- 
ley, the latter being loose on the sleeve, A. The 



band, B, may be bound to, or released from, the 
boss, C, by the action of right and left-handed 
screws controlled by the sliding sleeve, D, and lev- 
ers, E. It is evident that the band, B, moves with 





^ 


/'\ 


.4. 






\\ 




K^ ' 




♦T^l* E' ><T >^-L '^ i<- L ^ T >i* 

Fig. 193 — Spiral-jaw Detachable Coupling. 




Pig. 194— Band Friction Clutch. 

the shaft, and that the wheel or pulley will only 
rotate when the band, B, is tight on the boss, C. 
When used as a shaft coupling the boss, C, is pro- 



106 



MECHANICAL DRAWING 



longed and keyed to one of tlie shafts, the sleeve, 
A, being keyed to the other. 

Cone Friction Clutch. The friction clutch, 
shown in Fig, 195, although simpler in construc- 
tion than those which have already been described, 
is open to the objection that it requires a much 
greater force to put it into gear, and also this 
force, acting parallel to the axis of the shaft, 




Fig. 195— Cone Friction Clutch. 

causes an objectionable end thrust. The mean di- 
ameter of the conical part may be from four to 
eight times the diameter of the shaft, being larger 
the greater the amount of power which the clutch 
has to transmit. The inclination a of the slant 
side of the cone to its axis may vary from four de- 
grees to ten degrees. The other proportions may 
be as follows: 



B=2 D+1. E=.4 D-1-.4 

C=1.5D. T=.3D-|-.3. 

F=1.8D. t=.2D+.l. 

H=.5 D. 

Transmission of Motion by Belts. If motion be 
transmitted from one pulley to another by means 
of a thin inextensible belt as in Fig. 196, every 
part of the latter will have the same velocity, and 
the outer surface of the rim of each pulley will 




Fig. 196 — ^Belt Transmission of Power. 

haye the same velocity as the belt. TSiis fact en- 
ables us to prove the formula connecting the 
speeds of two pulleys with their diameters. Let 
Di and D2 be the diameters of the driver and fol- 
lower respectively, and let Ni and N, denote fceir 
speeds in revolutions per minute. The speed of 
the outer surface of the rim of the driver;=DiX 
3.1416 XNi, and the speed of the outer surface of 
the rim of the follower=D2X3.1416-l-N2, but each 
of these is equal to the speed of the belt, therefore, 
DiX 3.1416 xNi=D2+3.1416+N2; that is, D^^^= 

D2N2 or — = — . This formula is true whether the 
Ni D2 

belt is open or crossed, but the direction of the 



MACHINE DESIGN 



109 



motion will not be the same in each of these cases, 
as is shown by the arrows. With an open belt the 
pulleys rotate in the same direction, while with a 
cross belt, the pulleys rotate in opposite directions. 
Example 1. The driving pulley is 3 feet 6 inches 
in diameter, and it makes 100 revolutions per min- 
ute. The driver pulley is 2 feet 6 inches in diame- 
ter. To find its speed. 

42X100=30XN^ 
42X100 



therefore N2= 



30 



=140 rev. per minute. 



Example 2. A belt moving with a velocity of 
1000 feet per minute passes over a pulley 3 feet 3 
inches in diameter. To find the number of revolu- 
tions made by the pulley in one minute. 



Circumferential speed of pulle3r= 



39X3.1416XN 



12 



=1000, 



1000X12 
therefore 'N=;—ry^rrrTT^=97 .M revolutions per minute. 

oyAo.l4iD 

Split Pulleys. For convenience in fixing pulleys 
on shafts it is a common practice to make them in 
halves, which are bolted together as shown in Fig. 
197, which illustrates an example of a cast-iron 
split pulley 20 inches by 6 inches for a 2-inch shaft. 
By leaving a small clearance space between the 
two halves of the. pulley at the hub, the latter may 
be tightened on to the shaft so firmly by means 
of the bolts that no key is necessary. 



Wrought-Iron Pulleys. A' great number of 
wrought-iron pulleys are now made, and they pos- 
sess several important advantages over cast-iron 
ones. They are much lighter, and, being built up, 
they are free from the initial strains which exist 
in cast-iron pulleys due to unequal contraction in 
cooling. They are also safer at high speeds, not 




Fig. 197— Split Pulley. 

only because wrought-iron has a greater tenacity 
than cast-iron, but because if a wrought-iron pul- 
ley should burst, it will not fly to pieces like a cast- 
iron one, since cast-iron is so much more brittle 
than wrought-iron. Wrought-iron pulleys can be 
made of any diameter required, but it is when they 
are of large diameter that their superiority over 
cast-iron pulleys is most decided. 



110 



MECHANICAL DRAWING 



In a wrought-iron pulley the rim is made of 
sheet-iron, and riveted to the arms, which are 
made from bar-iron. The hub is sometimes also 
made of wrought-ixon, but generally it is made of 
cast-iron. 

The differences in the design of wrought-iron 
pulleys are chiefly in the form of the arms. 

Crank-Disks. In small engines, and in engines 
which run at a high speed, a cast-iron disk fre- 




rig. 198— Crank Disks, 

quently takes the place of the ordinary crank. 
This disk is hollowed out on the side next the 
crank-pin, as shown in Fig. 198, and the extra 
weight on the opposite side balances the crank -pin, 
and the part of the connecting-rod which revolves 
with it. Two crank-disks may be used to take the 
place of a double-armed crank. 

The following rules may be used in proportion- 
ing crank-disks: 



Di=D to 1.2D. 
B =.7D to D. 
A=.9B. 
C=.25D. 



F=2d. 

H=d to 1.5d. 
I^=B to 1.25B. 



A Stuffing-Box is used where a sliding or rotat- 
ing piece passes through the end or side of a ves- 
sel containing a fluid under pressure. The stuffing- 
box allows, the sliding or rotating piece to move 





"nl^. 



Mr *-*T 



m 



d. 







Fig. 199— Piston Ilod Stuffing-box. 



MACHINE DESIGN 



Hi 



freely, without allowing any leakage of the fluid. 
Applications of the stuflfing-box are found in the 
steam-engine where the piston-rod passes through 
the cylinder cover, and where the valve spindle 
passes through the valve casing; also at the trun- 
nions of the cylinder in an oscillating engine, and 
where the shaft of a centrifugal pump passes 
through the pump casing. Stuffing-boxes are also 
used to permit of the expansion and contraction of 
steam pipes. 

Fig. 199 shows an ordinary form of stuffing-box 
for the piston-rod of a vertical engine. AB is the 
piston-rod, CD a portion of the cylinder cover, and 
EF the stuffing-box. Fitting into the bottom of 
the stuffing-box is a brass bush, H. The space 
around the rod AB is filled with packing, of which 
there are a great many kinds, the simplest being 
greased hempen rope. The packing is compressed 
by screwing down the gland LM by means of the 
bolts P. When more than two bolts are used, the 
gland flange is generally made circular as shown 
in the lower left-hand view. For two bolts the 
gland flange is of the form shown in the lower 
right-hand view. When the gland is made of cast- 
iron it is generally lined with a brass bushing. 
When the gland is of brass no lines are necessary. 

Steam Cock. Fig. 200 shows a good example of 
a cock, suitable for steam or water, and adjacent 
to it is a table of dimensions, taken from actual 
practice, for four different sizes. The two parts of 




Fig. 200— Steam Cock. 



112 



MECHANICAL DEAWING 



the pipe meeting at the plug may be in the same 
straight line, or one may proceed from the bottom 
of the casing, as shown by the dotted lines. 



A 


1% 


2 


2% 


2X 


N 


1 3 


IX 


1 5 


1% 


B 


7>i 


7% 


8X 


9 





3 


3% 


Ql 3 


4A 


C 


5 


5% 


6 


6% 


P 


2tt 


2% 


nl 1 


3 


D 


1% 


2A 


2% 


2% 


Q 


3X 


4 


4X 


5 


E 


2tV 


2% 


211 


3tV 


E 


lA 


IX 


Hi 


113 
■l-TT 


F 


% 


1 


1 


1% 


S 


% 


% 


3/ 


7 


G 


1% 


lA 


IX 


lA 


T 


% 


% 


7 
TTT 


tV 


H 


4'A 


5>^ 


5% 


6 


T, 


t'V 


5 
TTT 


% 


% 


I 


% 


1 7 


tV 


% 


T. 


7 


7 
TTT 


X 


X 


J 


% 


1 7 


A 


% 


U 
V 


X 


% 


% 


% 


Ji 


7 
TTT 


1 5 
¥"2- 


K 


9 
TTT 


X 


X 


X 


% 


K 


2% 


8X 


3% 


311 


w 


3% 


3% 


3X 


4% 


L 


1% 


1% 


ItV 


1% 


X 


2X 


2tV 


2% 


3X 


M 


1% 


2 


9 3 


2X 













Shaft Hanger with Adjustable Bearing. Figui-e 
201 shows an excellent form of hanger and bear- 
ing for a shaft. The bearing is very long, and is 
held at the center of its length on spherical seats, 
which are formed on the ends of the vertical ad- 
justing screws. The spherical seats permit of a 
slight angular movement of the bearing, a move- 
ment which is necessary "with such a long bearing, 
to ensure that the axis of the bearing coincides 
with the axis of the shaft. The bearing and the 
vertical adjusting screws are made of cast-iron. 

Locomotive Connecting-Rod End. A modified 
form of box end is shown in Fig. 202. This is the 
design adopted on some roads for the large or 



crank-pin ends of the connecting-rods. The end 
of the rod is forked, and the open end is closed by 
a block which is held in position by a bolt. This 
bolt has a slight taper, and is screwed at both ends 
and provided with nuts, so that it may be easily 
tightened up, and also easily withdrawn. The 




tNLARGEWElNT 




SECTION AT A.B. 




Pig. 201 — Shaft-hanger with Adjustable Bearing. 



MACHINE DESIGN 



113 



brasses, -which have flanges on each side, are tight- 
ened up by a wedge-shaped block attached by two 
pins to a bolt screwed at both ends. 




Fig. 202 — ^Locomotive Connecting-rod End. 

Since the angular motion of a long connecting- 
rod is small, the wear of the brasses at the cross- 
head end is slight. For this reason the bearing 



at the cross-head end of locomotive connecting- 
rods is sometimes made solid. 

Locomotive Cross-Head. Fig. 203 shows a form 
of cross-head used on certain railroads. In this 
form the cross-head and slide-blocks are made in 
one piece of cast-steel. The cross-head pin is 
tapered, and carries a phosphor bronze bushing or 




Pig. 203 — Locomotive Cross-head. 



sleeve at its center, upon which works the connect- 
ing-rod end, the latter being solid and fitted with 
a hardened steel bushing. The phosphor bronze 
bushing is prevented from rotating on the cross- 
head pin by a sunk key, which also prevents the 
pin from rotating. 

Horsepower Transmitted by Ropes. Let Tj be 
the tension on the tight or driving side of a rope, 



114 



MECHANICAL DRAWING 



and Tg the tension on the slack side. The driving, 
force is then Tj— T2=P. Let HP be the horsepow- 
er transmitted by one rope, and V the velocity of 
the rope in feet per minute, then 



HP= 



VP 
33,000 



and P= 



33,000HP 



The following table has been calculated on the 
assumption that ^=4, and that the greatest stress 

■••2 

on the rope is 140 pounds per square inch of the 
cross section of the rope, so that the driving force 
P=Ti — T2, is 105 pounds per square inch of the 
cross section of the rope. The formula for the 
horsepower for one rope then becomes 



HP= 



D^X. 7854 x105V D'V 



33,000 



400 



very nearly^ where D is the diameter of the rope. 
If C, the circumference of the rope, be used instead 
of the diameter, then 



HP = 



3949.44 



or very nearly 



4000" 



Diam. 

of 
Rope 

in 
Inches. 


Circmn. 
of Rope 

in 
Inches. 


Horse Power Transmitted by one Rope when the Speed 
in Feet per Minute is: 


3000 


3500' 


4000 


4500 


5000 


5500 


6000 


1 

li 

li 

11 

li 

If 

If 

11 

2 


3.14 
3.53 
3.93 
4.32 
4.71 
5.10 
5.50 
5.89 
6.28 


7.50 
9.48 
11.72 
14.18 
16.87 
19.80 
22.97 
26.37 
30.00 


8.75 
11.07 
13.67 
16.54 
19.69 
23.11 
26,80 
30.76 
35.00 


30.00 
12.66 
15.62 
18.91 
22.50 
26.41 
30.62 
35.15 
40.00 


11.25 
14.24 
17.58 
21.27 
25.31 
29.71 
34.45 
39.55 
45.00 


12.50 
15.82 
19.53 
23.63 
28.12 
33.01 
38.28 
43.95 
50.00 


13.75 
17.40 
21.48 
26.00 
30.94 
36.31 
42.11 
48.34 
55.00 


15.00 
18.98 
23.44 
28.36 
33.75 
39.61 
45.94 
52.73 
60.00 



HORSEPOWER OF GEARS 



1. When the circular pitch is given — to find the 
horsepower capable of being transmitted by cast- 
iron gears with cut teeth: Multiply the pitch di- 
ameter of the gear by the circular pitch of the 
teeth, by the width of the teeth (all in inch meas- 
urements), and by the number of revolutions of 
the gear per minute. Divide the product by 550 
and the result will be the horsepower the gear is 
capable of transmitting. 

Let D be the pitch diameter of the gear, C the 
circular pitch and E the width of the tooth (all in 
inch measurements), R the number of revolutions 
of the gear per minute and H. P. the horsepower 
the gear is capable of transmitting, then 



HP = 



DXCXFXR 
550 



(1.) 



Example: What horsepower will the following 
cast-iron gear with cut teeth transmit at 100 revo- 
lutions per minute? The circular pitch of the gear 
is 2 inches, the number of teeth 33 and the width 
of the face of the tooth 2 inches. 

Answer: As the pitch diameter of the gear is 
approximately 21 inches, then 

21x2x2x100 



HP=: 



550 



: = 15.27 



Note: A cast-iron gear with cut teeth of 1 inch 
circular pitch and 1.048 inches width of tooth and 
with 33 teeth will transmit 1 horsepower at 50 rev- 
olutions per minute. As the pitch diameter of the 
gear is approximately 10^^ inches, then 



10.5X1X1.048X50 
550 



= 1 horsepower. 



2. When the diameter pitch is given to find the 
horsepower capable of being transmitted by cast- 
iron gears with cut teeth: Multiply the pitch diam- 
eter of the gear by the width of the tooth (both 
in inch measurements), and by the number of revo- 
lutions of the gear per minute. Divide the product 
by the Diametral pitch and by 175, and the result 
will be the horsepower the gear is capable of trans- 
mitting. 

Let D be the pitch diameter of the gear, F the 
width of the tooth (both in inch measurements), 
R the number of revolutions of the gear per min- 
ute of the gear, P the diametral pitch and H. P. 
the horsepower, then 

DxFxR 



HP=- 



(2) 



115 



PX175 

Example: What horsepower will the following 
cast-iron gear with cut teeth transmit at 100 revo- 



116 



MECHANICAL DRAWING 



lutions per minute? The diametral pitch, of the 
gear is 1%, the width of the face of the tooth 2 
inches, and the pitch diameter 20 inches. 

Answer: 20x2x100 ^co^t. 

— -^ =15.24 horsepower. 

3. To find the horsepower capable of being 
transmitted by a gear with cut teeth of any given 
material. Multiply the results obtained by Eule 1 
or 2, or by Formula 1 or 2, by the coefficients for 
the various metals given herewith. 

Cast iron being taken as 1 or unity, then: Malle- 
able Iron=1.25, Brass=1.33, Bronze=1.66, Gun 
Metal=2.00, Phosphor Bronze=3.00, Wrought- 
Iron=3.33, Steel=4.00. 

Example: If a cast-iron gear of given dimen- 
sions will transmit 2 horsepower, what horsepower 
will a similar gear if made of phosphor bronze? 

Answer: As the coefficient for phosphor bronze 
is 3, then 2x3=6 horsepower that the gear will 
transmit if made of phosphor bronze in place of 
cast iron. 

Note: If the diametral instead of the circular 
pitch be given. To iind the circular pitch of the 
teeth, divide 3.1416 by the diametral pitch of 
the gear. 

Example: Required the circular pitch of the 
teeth of a gear of 4 diametral pitch? 

Answer: 3.1416 divided by 4, gives .7854 as 
the circular pitch in inches of the gear teeth. 



Example: What is the circular pitch of a gear 
of 2 diametral pitch ? 

Answer: 3.1416 divided by 2, gives 1.5708 
inches as the circular pitch of the gear teeth. 

Transmission of Motion by Gears. 

Motion is in many cases transmitted by means 
of gear wheels, and accordingly as the driving and 
driven are of equal or unequal diameters, so are 
equal or unequal velocities produced. 

When time is not taken into account. Divide 
the greater diameter, or number of teeth, by the 
lesser diameter, or number of teeth, and the 
quotient is the number of revolutions the lesser 
will make for 1 of the greater. 

Example: How many revolutions will a pinion 
of 20 teeth malie for 1 of a gear with 125 teeth? 

Answer: 125-^20=6.25, or 6^4 revolutions. 

Intermediate geajrs of any diameter, used to con- 
nect other gears at any required distance apart, 
cause no variation of velocity more than other- 
wise would result if the first and last gears were 
in mesh. 

To find the number of revolutions of the last, 
to 1 of the first, in a train of gears and pinions. 
Divide the product of all the teeth in the driving 
by the product of all the teeth in the driven gears, 
and the quotient will equal the ratio of velocity 
required. 



HORSEPOWEE OF GEARS 



117 



Example: A gear of 42 teeth giving motion to 
one of 12 teeth, on which shaft is a pulley of 21 
inches diameter, driving one of 6 inches diameter, 
required the number of revolutions of the last 
pulley to one of the first gear. 

Answer: (42x21)-^(12x6)=12.25, or 12% 
revolutions. 

Where increase or decrease of velocity is re- 
quired to be communicated by gears, it has been 
demonstrated that the number of teeth on the 
pinion should not be less than 1 to 6 of its wheel, 
unless there be other reasons for a higher ratio. 

When time must be regarded. Multiply the 
diameter, or number of teeth in the driving gear, 
by its velocity in any given time, and divide the 
product by the required velocity of the driven 
gear, the quotient equals the number of teeth, or 
diameter of the driven gear, to produce the veloc- 
ity required. 

Example : If a gear containing 84 teeth makes 
20 revolutions per minute, how many teeth must 
another contain to work in contact, and make 60 
revolutions in the same time? 

Answer: (84x20) --60=28 teeth. 

The distance between the centers and velocities 
of two gears being given, to find their proper 
diameters. Divide the greatest velocity by the 
least. The quotient is the ratio of diameter the 
wheels must bear to each other. Hence, divide 
the distance between the centers by the ratio plus 



1. The quotient will equal the radius of the 
smaller gear, and subtract the radius thus ob- 
tained from the distance between the centers, the 
remainder will equal the radius of the other gear. 
Example: The distance of two shafts from cen- 
ter to center is 50 inches, and the velocity of one 
shaft is 25 revolutions per minute, the other shaft 
is to make 80 revolutions in the same time. Ee- 
quired the proper diameters of the gears at the 
pitch lines. 

Answer: 80^-25=3.2, the ratio of velocity, and 
50-f-(3.2-t-l)=11.9, the radius of the smaller 
wheel; then 50— (11.9x38.1) the radius of the 
larger gear. Their diameters are therefore 
11.9X2=23.8, and 38.1x2=76.2 inches. 

To obtain or diminish an accumulated velocity 
by means of gears and pinions, or gears, pinions, 
and pulleys, it is necessary that a proportional 
ratio of velocity should exist, and which is ob- 
tained thus: Multiply the given and required 
velocities together, and the square root of the 
product is the mean or proportionate velocity. 

Example: Let the given velocity of a gear con- 
taining 54 teeth equal 16 revolutions per minute, 
and the given diameter of an intermediate pulley 
equal 25 inches, to obtain a velocity of 81 revolu- 
tions in a machine. Eequired the number of teeth 
in the intermediate gear, and the diameter of the 
last pulley. 



118 



MECHANICAL DRAWING 



Answer: V81x 16=36 the mean velocity, 
(54X16)^36=24 teeth, and (25x36)^81=11.1 
inches, the diameter of the pulley. 

Diametral Pitch System of Gears. 

The Diametral pitch system is based on the 
number of teeth to one inch diameter of the pitch 
circle. Formulas are herewith given so that if 
the number of teeth in the gear and the diametral 
pitch are known, the pitch diameter of the gear 
may be found, also the outside diameter, the work- 
ing depth and clearance at the bottom of the 
tooth. Let P be the pitch diameter in inches, D 
the diametral pitch of the gear, C the circular 
pitch in inches^ the outside diameter in inches, 
T the thickness of the tooth at the pitch line in 
inches, W the working depth of the tooth in 
inches, and N the number of teeth in the gear, 
then 



(1.) 
(2.) 
(3.) 
(4.) 



W=Working depth of tooth =^=2-=-D (5.) 



N= Number of teeUi 
T = Thickness of tooth 



= PXD (6.) 
= 1.571-=-D (7.) 



P= Pitch diameter 


N 
D 


= Outside diameter 


-4 


D = Diametral pitch 


N 
~P 


= Circular pitch 


3.142 



Clearance at bottom of tooth = 



0.157 
D 



(8.) 



Example: Kequired, the pitch diameter of a 
gear with 20 teeth and 4 diametral pitch. 

Answer: From Formula 1, as the pitch diam- 
eter is equal to the number of teeth divided by the 
diametral pitch, then 20 divided by 4 equals 5, 
as the required pitch diameter in inches. 

Example: What is the outside diameter of the 
same gear? 

Answer: From Formula 2, as the pitch diam- 
eter is 5 inches and the diametral pitch 4, then 4 
plus 2-4 equals 4% as the proper outside diameter 
for the gear. 

Example: "What should be the diametral pitch 
of a. gear with 30 teeth and 6 inches pitch diam- 
eter? 

Answer: From Formula 3, 30 divided by 6 
equals 5, as the diametral pitch to be used for the 
gear. 

Example: Required the circular pitch of the 
teeth of a gear whose diametral pitch is 6. 

Answer: From Formula 4, 3,142 divided by 6 
gives 0.524 inches as the circular pitch of the teeth 
of the gear. 

Example: What should be the working depth 
of a tooth of 4 diametral pitch? 



HORSEPOWER OP GEARS 



119 



Answer: From Formula 5, 2 divided by 4 gives 
0.5 or one-half an inch as the working depth of 
the tooth. 

'Example: How many teeth are there in a gear 
of 7 inches pitch diameter and 7 diametral pitch? 

Answer: From Formula 6 the number of teeth 
is equal to 7 multiplied by 7, or 49 teeth in the 
gear. 

Example: What is the thickness at the* pitch 
line of a tooth of 8 diametral pitch? 




Fig. 204— Rack with Straight Face Involute Teeth. 

Answer: By Formula 7 the thickness of the 
tooth at the pitch line is 1.571 divided by the 
diametral pitch, then 1.571-^8 gives 0.196 inches 
as the thickness of the tooth. 

Example: What should be the correct clearance 
at the bottom of a tooth of 3 diametral pitch ? 

Answer: From Formula 8 the clearance at the 
bottom of the tooth is equal to 0.157 divided by 3, 
which gives 0.052 as the required clearance. 



When the pitch circle becomes of infinite diam- 
eter, as in a rack, the base circle will also become 
of infinite diameter, and the involute will become 
a straight line. Hence, in a rack which gears with 
a wheel having involute teeth, the teeth are 
straight on face and flank, as shown in Fig. 204. 
The faces and flanks are at right angles to the path 
of contact, and therefore make an angle of 75 
degrees with the pitch line. 

Table No. 10 gives the dimensions of Involute 
Tooth Spur Gears from 1 to 16 Diametral pitch. 



Dimensions op 


Involute Tooth 


Spur Gears. 


Diametral 
Pitch. 


* Circular 
Pitch. 


Width of 
Tooth on 


Working 
Depth of 


Actual 
Depth of 


Clearance 
at Bottom 


Pitch Line. 


Tooth 


Tooth. 


of Tooth. 


1 


3.142 


1.571 


2.000 


2.157 


0.157 


2 


1.571 


0.785 


1.000 


1.078 


0.078 


3 


1.047 


0.524 


0.667 


0.719 


0.052 


4 


0.785 


0.393 


0.500 


0.539 


0.039 


5 


0.628 


0.314 


0.400 


0.431 


0.031 


6 


0.524 


0.262 


0.333 


0.360 


0.026 


7 


0.447 


0.224 


0.286 


0.308 


0.022 


8 


0.393 


0.196 


0.250 


0.270 


0.019 


10 


0.314 


0.157 


0.200 


0.216 


0.016 


12 


0.262 


0.131 


0.167 


0.]80 


0.013 


14 


0.224 


0.112 


0.143 


0.154 


0.011 


16 


0.196 


0.098 


0.125 


0.135 


0.009 



♦The circular pitch corresponding to any diametral pitch number, may 
be found by dividing the constant 3.1416 by the diametral pitch. 

Example: What is the circular pitch in inches corresponding to 4 di- 
ametral pitch? 

Answer: Dividing 3.1416 by 4 gives 0.7854 inches as the required cir- 
cular pitch. 



120 



MECHANICAL DEAWING 



Worm Gearing. A screw wliicli gears with the 
teeth of a wheel is called a worm, and the wheel 
is called a worm wheel. In worm gearing the axis 
of the worm is usually at right angles to the axis 
of the wheel, but this is not absolutely necessary. 

If a section of the worm and wheel be taken by 
a plane containing the axis of the worm, the teeth 
of the wheel and the threads of the worm in this 



section should be the same as for an ordinary 
toothed wheel and a rack. The curves of the sec- 
tion of the teeth may be cycloidal, but since the 
teeth of a rack to gear with a wheel with involute 
teeth are straight from root to point, it is better 
to make the teeth of a worm wheel of the involute 
shape, because then the tool for cutting the worm 
is of a very simple form. 



STEAM BOILERS 



In designing a steam boiler for a given engine, 
the probable indicator diagram is first drawn, and 
from this the volume of steam used, at a partic- 
ular pressure, in one stroke is determined. From 
a table of the properties of steam which is usually- 
given in treatises on the steam-engine, the weight 
of a cubic foot of this steam may be found, and 
from this the weight of steam used in one stroke 
is calculated. Multiplying the weight of steam 
used in one stroke by the number of strokes per 
hour, the weight of steam used per hour is deter- 
mined. 



Properties of Saturated Steam 






Total Units of 

Heat in 1 lb. 

weight of 

Steam 


0) 


£S4 
5-5 S 

<1 CO 




Total Units of 

Heat in 1 lb. 

weight of 

Steam. 


■o " 5 


14.7 


212.0 


1178.1 


.038 


85 


316.1 


1209.9 


.198 


25. 


240.1 


1186.6 


.062 


95 


324.1 


1212.3 


.219 


35. 


259.3 


1192.5 


.085 


110 


334.6 


1215.5 


.252 


45. 


274.4 


1197.1 


.108 


125 


344.2 


1218.4 


.286 


55. 


287.1 


1201.0 


.131 


145 


355.6 


1221.9 


.329 


65. 


298.0 


1204.3 


.153 


165 


366.0 


1224.9 


.371 


75. 


307.5 


1207.2 


.175 


185 


375.3 


1227.8 


.414 



In actual practice, the amount of water evapo- 
rated in the boiler is from 1.2 to 1.8 times the 
amount shown by the indicator diagram of the 
engine. 

Example: To find the weight of water which 
must be evaporated per hour by a boiler to supply 
steam to an engine having a cylinder 10 inches in 
diameter, with a piston stroke of 18 inches. Steam 
cut off at one-third of the stroke. Revolutions per 
minute 100. Pressure of steam 50 pounds per 
square inch by gauge, or 65 pounds per square 
inch absolute. 



Volume of steam used in one stroke 
102 X. 7854x18 



3x1728 



cubic feet. 



Weight of one cubic foot of steam at an abso- 
lute pressure of 65 pounds per square inch, from 
the Table=.153 pound. 

Weight of steam used in one stroke, 



102X.7854xl8x.153 
3x1728 



pounds. 



Weight of steam used in one minute, 
102x.7854xl8x. 153X2X100 



121 



3X1728 



pounds. 



122 



MECHANICAL DRAWING 



Weight of steam used in one hour, 

10" X. 7854 X 18 X. 153X2X100X60 
3X1728 



pounds. 



= 503.3 pounds. 

In addition to the above quantity of steam, 
there will be that required to fill the clearance 
spaces in the cylinder, also a quantity to make up 
for the loss due to condensation in the cylinder 
*and steam pipe. Allowing a margin of 40 per 
cent, the boiler will require to evaporate 503.3 X 
1.4=704.6 pounds, say 705 pounds of water per 
hour under the given pressure. 

Equivalent Evaporation from and at 212°. For 
the purpose of comparison it is usual to state the 
weight of steam produced in a boiler on the as- 
sumption that the feed water is supplied at a tem- 
perature of 212° F., and that it is evaporated at 
this temperature. 

Let W = weight of steam produced at any given temper- 
ature or pressure. 
H = total heat in 1 pound weight of this steam. 
t = temperature of feed water. 
' Wi = equivalent weight of steam at 212 degrees tem- 
perature produced from feed water at 212 
degrees. 
Then Wi(1178.1-212) =W(H-0 

, ,^ W(H-0 

Example: To determine the equivalent evapo- 
ration from and at 212° in the boiler for the engine 



in the example worked out in the preceding ar- 
ticle, supposing that the feed water is supplied 
at a temperature of 60°. 



W,= 



705(1204.3-60) 705x1144.3 



966.1 



966.1 



= 835 pounds. 



Evaporative Performances of Steam Boilers. 

Theoretically 1 pound of coal should evaporate 
from 12 to 16 pounds of water from and at 212°, 
the amount depending on the quality of the coal. 
The following table gives the weight of water 
evaporated from and at 212° by 1 pound of aver- 
age coal in various types of boilers: 



Type of Boiler 


Water evaporated 

from and at 212" 
per pound of Coal. 


Vertical cross-tube boilers 


Pounds. 

5 to 7 

6 " 9 

7 " 9 

8 " lOi 
8 " 10 
8 " 10 
8i " 11 
7 " 8i 


Vertical multitubular boilers 


Cornish boilers 


Lancashire boilers 


Sectional or water-tube boilers 


Marine boilers 


Locomotive boilers 


Torpedo-boat boilers 





Vertical Cross Tube Boilers. Vertical boilers 
possess the advantage of taking up a compara- 
tively small amount of floor area. In its simplest 
form the vertical boiler consists of a cylindrical 
shell surrounding a nearly cylindrical fire-box, in 
the bottom of which is the grate. A tube, called 
an uptake, passes from the crown of the fire-box 



STEAM BOILERS 



123 



to the crown of the shell, where it is connected 
to a chimney. To increase the amount of heating 
surface and improve the circulation of the water, 
and also to increase the strength of the fire-box, 
the latter is fitted with one or more cross-tubes as 
shown in Fig. 205. 

The vertical seam of the fire-box shell may be 
a single riveted joint, but in the best work the 
joint is lap welded. The fire-box shell slopes to 
the vertical at the rate of % inch to % inch per 
foot of length. 

The cross-tubes are either flanged and riveted to 
the fire-box or they are welded to it. With welded 
joints in the fire-box double thicknesses of plates 
are avoided and the chances of overheating at the 
joints reduced. When the cross tubes are welded 
in, however, it is more difficult to renew them if 
they should wear out before the fire-box. 

The cross tubes are placed slightly inclined to 
ensure a more efficient circulation of the water. 

The uptake is connected to the crown of the fire- 
box and to the crown of the outside shell by 
flanged joints as shown in Fig. 205, or by welded 
angle iron rings which are riveted to the crowns 
and to the uptake. 

The part of the uptake which passes through the 
steam space is very liable to become overheated, 
and to prevent this it is enlarged in diameter from 
the top to the low-water level and lined with fire- 
clay. A cast-iron liner is sometimes introduced 




Fig. 205— Vertical Cross Tube Boiler. 



124 



MECHANICAL DRAWING 



for the same purpose. The cast-iron liner may fit 
close into the uptake, or it may be separated from 




it all round by an air space not exceeding % inch 
wide as shown in Fig. 205. 

Semi-Portable Boiler. Boilers of the locomotive 
type are now largely used for supplying steam to 
engines for many purposes on land, and they are 
also used in torpedo boats. Boilers of this class, 
when used in a fixed position on land, are some- 
times called semi-portable, and sometimes semi- 
fixed. Fig. 206 shows a longitudinal section of the 
boiler. The shell plates are of steel, and the inter- 
nal fire-box of steel of special quality. The plates 
are flanged by hydraulic machinery, and angle- 
irons are dispensed with, except at the front end 
of the smoke-box, and there a flanged ring is used. 

The edges of the plates are planed, and the outer 
face of the flange of the smoke-box tube plate is 
turned. All the rivet-holes are drilled after the 
plates have been bent in position, and the riveting 
is done by machinery. 

The circumferential seams of the barrel are 
double riveted, with rivets % inch diameter at 3^/4 
inches pitch. The width of the lap is SVs inches, 
and the distance between the center lines of the 
two rows of rivets is 1% inches. 

The longitudinal seams of the barrel are double 
riveted, butt joints, strapped outside and inside. 
The rivets are % inch diameter at 3 3-16 inches 
pitch. The distance between the center lines of 
the two inner of the four rows of rivets is 3 inches. 
The distance between the center line of each outer 



STEAM BOILERS 



125 



row and the center line of the adjacent inner row 
is lYs inches. The total width of the butt straps 
is 9% inches. 

The tubes have an external diameter of 2^/2 
inches, and they are % inch thick. 

The smoke-box tube plate and the back end 
plate are stayed to the shell by long gusset stays, 
and they are also stiffened by T-irons. The sides 
of the inner fire-box are stayed to the outer fire- 
box by screwed stays. The crown of the inner 
fire-box is stayed by girder stays, which are placed 
transversely as shown. 

The boiler is supported at the fire-box end on a 
cast-iron frame, not shown in the illustrations, 
which serves as an ash-pit. The smoke-box end is 
supported on a hollow cast-iron pillar, into which 
the exhaust steam from the engine is led on its 
way to the chimney. 

There is a total heating surface of 290.5 square 
feet, and a grate area of 11.2 square feet. 

The working steam-pressure is 140 pounds per 
square inch. 

This boiler, which is a 20 nominal horsepower 
boiler, is capable of supplying steam to a com- 
pound engine having high and low pressure cylin- 
ders of 9 inches and 14 inches diameter respec- 
tively, with a piston stroke of 16 inches, the crank- 
shaft making 135 revolutions per minute. 

Gusset Stays. A gusset stay is a diagonal stay 




Pig. 207— Qusset Stays. 



126 



MECHANICAL DRAWING 



in which a flat plate is used instead of a bar or 
rod. 

The most important application of the gusset 
stay is to be found in Cornish and Lancashire 
boilers for staying the flat ends to the cylindrical 
shell. 

The following description of the arrangement 
of the gusset stays in a Lancashire boiler, 7 feet 
6 inches in diameter, for a working pressure of 
100 pounds per square inch, is given: 

To arrange the staying so that it shall give ade- 
quate support to the flat ends without rendering 
them unnecessarily rigid, is a point of consider- 
able importance in boilers of the single or double 
flued internally fired class. If the ends be made 
too stiff, there is a serious risk of grooving being 
set up, either in the end plate at the edge of the 
upper part of the furnace tube angle iron, or at 
the root of the angle iron itself. This is more 
particularly the case at the front end, and arises 
from the varying thrust of the upper portion of the 
furnace tube upon the end plate. To avoid the 
risk of grooving, therefore, a certain space should 
be allowed between the rivets of the furnace tube 
angle iron and the lowest rivets of the gussets. 



This is a point in which a mistake is often made, 
and, with a desire to render the end plate suffi- 
ciently strong, the gussets are extended until they 
almost touch the furnace tube angle iron, the result 
being that a grooving action is set up when the 
boiler is put to work, and, although the defect un- 
der such circumstances may not be a source of 
danger, yet it gives rise to considerable annoy- 
ance and inconvenience, and is sure sooner or later 
to necessitate extensive repairs. 

The arrangement of staying shown in Fig. 207 
is the outcome of extensive experience, and repre- 
sents, in fact, the most modern practice of the 
best makers of this class of boiler. 

The arrangement of the gussets, it will be seen, 
follows a simple geometrical arrangement. The 
gussets are pitched uniformly at a distance of 21 
inches apart, measured along the circumference 
of the shell, except at the bottom of the front end, 
where the pitch is reduced to 19 inches, with a 
view to bringing them as close to the mudhole 
mouthpiece as possible, and thus assist in support- 
ing the flat space below the tubes. At the back 
end, as there is no mudhole to interfere with the 
arrangement, a single central gusset stay is 
adopted. 



STEAM ENGINES 



To ascertain the horsepower when the average 
pressure upon the piston in pounds per square 
inch is known. 

Find the area of the piston in square inches, 
by multiplying the square of its diameter by 
0.7854. 

Find the total pressure in pounds on the piston 
by multiplying its area by the average effective 
pressure in pounds per square inch. The average 
effective pressure is the average pressure in 
pounds per square inch less 14.7, which must be 
deducted to allow for the atmospheric pressure 
against the piston. 

Find the useful piston travel in feet per minute 
by multiplying twice the length of the piston 
stroke in feet by the number of revolutions per 
minute of the crank shaft, for a double-acting 
steam engine. Find the energy in foot pounds per 
minute by multiplying the total pressure in pounds 
on the piston by the useful piston travel in feet per 
minute. 

The horsepower may then be ascertained by di- 
viding the energy in foot-pounds per minute by 
33,000. 

While there are numerous formulas in use for 



127 



calculating the horsepower of an engine, one of the 
most simple is as follows: 

PXLXAXN 
33,000 

Where P is the average effective pressure in 
pounds per square inch, L twice the length of the 
piston stroke in feet, A the area of the piston 
in square inches and N the number of revolutions 
of the crank shaft per minute. 

Length of Stroke of Piston. If the speed of the 
piston in feet per minute S, and the number of rev- 
olutions, R, of the crank shaft per minute are 

given, then the length of the stroke I in feet is 

g 
given by the formula Z= r^. 

No definite rule can be given for the length of 
the stroke. In single cylinder engines the stroke 
of the piston is generally made to depend on the 
diameter of the cylinder, but the relation is very 
variable, some engineers make the length of the 
stroke equal to twice the diameter of the cylinder. 
In locomotive engines the stroke of the pistons 
varies from 1.2 to 1.55 times the diameter of the 
cylinders. 

Speed of Piston. The speed of a piston is usual- 



128 



MECHANICAL DRAWING 



ly given in feet per minute, and is the distance in 
feet which it goes in one minute. This is the mean 
speed. The speed varies from nothing at the be- 
ginning and end of each stroke to a maximum near 
the middle of the stroke. The mean speed of the 
piston is determined by multiplying the length 
of the stroke in feet by the number of strokes per 
minute. 

In case of a locomotive engine, if ^^ength of 
stroke of pistons in feet, D=diameter of driving 
wheels in feet, M^speed of train in miles per hour, 
and S==speed of pistons in feet per minute, then 

M x5280x2x? 56.02M? 
DX 3.1416X60 ~ D 
For engines of the same class the speed of piston 
is generally greater, the greater the length of the 
stroke, and may be taken as proportional to the 
cube root of the stroke. 

The following table gives speeds of pistons to 
be met with in ordinary practice: 



Class of Engine. 


Speed of Piston 

in Feet per 

Minute. 


Ordinary direct-acting pumping engines (non- 
rotative) 


90 to 130 
200 to 400 

400 to 800 
400 to 650 
700 to 900 
800 to 1000 
1000 to 1200 


Ordinary horizontal engines 


Horizontal compound and triple-expansion 
mill engines 


Ordinary marine engines 


Engines for large high-speed steamships 

Locomotive engines 


Engines for torpedo-boats 





Clearance and Clearance Volume. The term 
clearance as applied to steam cylinders may mean 
the distance between the cylinder cover and the 
piston when the latter is nearest to the cover, or 
it may mean the volume of the space between the 
piston and the steam valve when that space is 
least — that is, when the piston is at the beginning 
of its stroke. 

Clearance in a steam cylinder is necessary to 
provide for any slight inaccuracy in the setting 
of the cylinder in relation to the crank shaft, 
to provide for inequalities on the surfaces of the 
piston and the cylinder cover, to provide for the 
slight errors which may occur in the lengths of 
the piston-rod, connecting-rod, and crank arms, 
and to provide for the wear which takes place at 
the cross-head, crank-pin, and crank-shaft bear- 
ings. 

The amount of the clearance varies with the size 
of the engine, being about three-eighths of an inch 
in small engines, and seven-eighths of an inch in 
large engines. In horizontal engines the clearance 
is generally the same at both ends of the cylinder, 
but in inverted cylinder engines, such as are now 
used in steamships, the clearance at the lower end 
is usually about one and a half times the clearance 
at the upper end. 

The clearance volume is usually expressed as a 
percentage of the volume swept through by the 



STEAM ENGINES 



129 



piston in one stroke, and it varies from 2 to 15 per 
cent in different cases. 

Classification of Valves. A valve is a piece of 
mechanism for controlling the magnitude or direc- 
tion of the motion of a fluid through a passage. 
The seat of a. valve is the surface against which 
it presses when closed, and the face of the valve 
is the portion of its surface which comes in con- 
tact with its seat. Valves may be classified accord- 
ing to the means by which they are moved. In 
one class the valves are moved by the pressure 
of the fluid. Nearly all pump valves belong to 
this class. The valves belonging to this class are 
automatic in their action, and they permit the 
fluid to pass in one direction only. Flap valves, 
flexible disc valves, lift valves (including rigid disc 
valves with flat seats, disc valves with bevelled 
edges fitting on conical seats, and ball valves), 
belong to the class of valves moved by the fluid. 
In another class the valves are moved by hand or 
by external mechanism, so that the motion of the 
valve is independent of the motion of the fluid. 

Since many of the valves belonging to the first 
class mentioned above may be converted into 
valves of the second class without altering the 
construction of the valves themselves, a better 
classification is as follows: flap valves, which bend 
or turn upon a hinge, lift valves, which rise per- 
pendicularly to the seat, sliding valves, which 
move parallel to the seat. 



Piston Valves. To overcome the friction of a 
slide valve of large area, working under a high 
steam pressure, requires the expenditure of a con- 
siderable amount of power. This waste of power 
may be reduced by fitting the back of the valve 
with a relief frame containing packing rings 
which press against the back of the valve casing, 
and so cut off a portion of the area of the valve 
from the pressure of the steam, but this and other 
arrangements for partly balancing the valve have 
not proved quite satisfactory. 

The great waste attending the use of large slide 
valves under high pressure steam has led engi- 
neers to adopt piston valves, which, although 
usually much more expensive to construct than 
ordinary slide valves, require very much less 
power to work them. 

A piston valve is really a slide valve in which 
the face and seat are complete cylindrical surfaces. 
A piston valve is perfectly balanced so far as the 
steam pressure is concerned, and the only friction 
to be overcome is that due to the small side 
pressure necessary to keep the valve steam tight. 
Piston valves are made steam-tight in much the 
same way as ordinary pistons, but since the 
amount of the motion of the valve is small com- 
pared with that of the engine piston, the wear is 
small, and the springs therefore may be stiff with 
very little range. 

Cylinder Flange. The thickness of the cylinder 



130 



MECHANICAL BRAWINO 




Fig. 208 — Simple Expansion, Piston-valve Engine, end 
Elevation. 



flange varies in practice from 1.2 to 1.4 times tlie 
thickness of the cylinder barrel. The width of 
the flange is determined by the size of the bolts 
for securing the cover to the cylinder. The dis- 
tance from the center of the bolts to the outside 
of the flange should not be less than ^-f i/4 inch, 
and need not be more than ly^c?, where d is the 
diameter of the bolt over the threads. 

Bolts for Cylinder Cover. The bolts used for 
securing the cylinder cover to the cylinder are 
nearly always stud bolts. Stud bolts are preferred 
to bolts with heads for two reasons : first, the cylin- 
der flange may be narrower when stud bolts are 
used, second, bolts with heads cannot be removed 
without interfering with the lagging round the 
cylinder. 

Let d = nominal diameter of bolts in inches; 

di = diameter of bolts at bottom of screw thread in 

inches. 
n = number of bolts. 
/ = stress on bolts in pounds per square inch of net 

section. 
D = diameter of cylinder. 

p = pressure of steam in pounds per square inch. 
Then .7854^?^/= .7854D> _ 

and(?i=D^^. 

Bolts of less than %-inch diameter should not 
be used for cylinder covers, and the stress/ should 
be less for small bolts than for large ones. / may 
be taken at 4000^?, but should not exceed 6000. 



STEAM ENGINES 



131 



The number of bolts depends on tbeir distance 
apart, which, again, depends on the thickness of 
the flange of the cover and the pressure of the 
steam. One rule gives the maximum pitch of the 

bolts equal to 40\/— , where t is the thickness of the flange 
in inches. 

Simple Expansion, Piston Valve Engine. Figs. 
208 and 209 show elevations of a simple expan- 
sion, piston valve steam engine. The main bear- 
ings are ring oiling, with a babbitted shell for 
the lower half which rests on a broad seat in the 
casting, thus giving firm support to the crank 
shaft and correct alignment. Encircling the shaft 
are bronze oil rings, running on the shaft mid- 
way of the bearings. Oil grooves in the babbitt 
shell are so cut that the oil can only get out of 
the bearings by working towards each end, whence 
it flows back to the reservoir in the center. 

Compound and Multiple Stage Expansion En- 
gines. The greater the amount of the expansion 
of the steam in a steam cylinder, the greater, the- 
oretically, is the amount of work obtained from a 
given weight of steam. But when steam expands 
its temperature falls, consequently the steam cylin- 
der is heated by the high pressure steam at the 
beginning of the stroke, a result which is followed 
by a partial condensation of that steam. Then, as 
the steam expands, its temperature falls, and so 
does that of the cylinder, part of the heat of the 




Fig. 209— Simple Expansion, Piston-valve Engine, Side 
Elevation. 



132 



MECHANICAL DEAWING 



latter going to warm the steam and re-evaporate 
the water deposited at tlie beginning of the stroke. 
It is, therefore, evident that with a great range 
of temperature in the cylinder, consequent on a 
high rate of expansion, a portion of steam may 
be condensed at the beginning of the stroke, and 
reconverted into steam towards the end of the 
stroke, and thus pass through the cylinder without 
doing work. 

To obtain the advantage following the high rate 
of expansion, and to obviate to some extent the 
loss due to the fall in temperature of the expanded 
steam, the compound engine was introduced. In 
the compound engine the steam is allowed to ex- 
pand to a certain extent in one cylinder, and is 
then passed into a larger cylinder and allowed to 
expand still further. In this way the range of 
temperature in each cylinder is much less than it 
would be if the whole expansion was carried out 
in one cylinder, and any condensed steam, re- 
evaporated at the end of the stroke in the first 
cylinder, does work in the second. But, neglecting 
the losses due to condensation and wire-drawing 
of the steam, the total work done in the two cylin- 
ders is the same as if the steam was used in the 
larger or low-pressure cylinder only, the total 
amount of expansion being the same in both cases. 

The range of temperature may be still further 
reduced by expanding the steam successively in 



three cylinders as in triple expansion engines, and 
still further by expanding successively in four 
cylinders as in quadruple expansion engines. 

Ratios of Cylinder Volumes in Compound and 
Multiple Stage Expansion Engines. "When steam 
is expanded in two or more cylinders successively, 
the considerations which determine the relative 
volumes of the cylinders are : The distribution of 
the power between the cylinders, the distribution 
of the initial loads on the pistons, and the range 
of temperature in each cylinder. When there is a 
separate crank for each cylinder it is very desir- 
able that the total power should, as nearly as pos- 
sible, be equally divided between the cylinders, 
and also that the initial or maximum loads on the 
pistons should be nearly equal. In all cases also 
the range of temperature should be as nearly as 
possible the same in each cylinder. 

No hard and fast rules can be given for the 
relative volumes of the cylinders of compound and 
multiple stage expansion engines. In actual prac- 
tice the ratios of the volumes of the cylinders vary 
considerably. The following rules and tables a-re 
based on the averages of a large number of exam- 
ples from recent practice, and may therefore be of 
service in determining, provisionally, the relative 
sizes of the cylinders. 

P=Pressure of steam in boiler by gauge in 
pounds per square inch. * 



STEAM ENGINES 



133 



Compound or two stage expansion condensing 
engines. 

Vol. of L.P. cylinder 4P+40 



Vol. ofH.P. cylinder 



100 



Tlie following examples have been worked out 
by the above rule, the volume of the high-pressure 
cylinder being taken as 1. 



Boiler Pressure by Gauge. 


60 


70 


80 


90 


100 


110 


120 

1. 
5.2 


Vol. of H.P. cylinder.... 
Vol. of L.P. cylinder .... 


1. 

2.8 


1. 

3.2 


1. 
3.6 


1. 
4. 


1. 

4.4 


1. 
4.8 



Triple or three stage expansion condensing 
engines. 

Vol. of I.P. cylinder P + lOO 



Vol. of H.P. cylinder 



100 



Vol. of L.P. cylinder 4P+50 



Vol. of H.P. cylinder 



100 



The following examples have been worked out 
by the preceding rules, the volume of the high- 
pressure cylinder being taken as 1. 



Boiler Pressure by Gauge. 


120 


130 


140 


150 


160 


170 


180 

1. 

2.8 

7.7 


Vol. of H.P. cylinder.... 
Vol. of I.P. cylinder .... 
Vol. of L.P. cylinder .... 


1. 

2.2 

5.3 


1. 
2.3 

5.7 


1. 

2.4 

6.1 


1. 

2.5 

6.5 


1. 

2.6 

6.9 


1. 

2.7 

7.3 



Jet Condensers — Quantity of Injection Water. 

The terminal pressure in the low-pressure cylinder 
of a steam engine may be taken at from 5 to 10 
pounds per square inch absolute, so that every 
pound weight of steam as it passes into the con- 
denser will contain from 1131 to 1140 British 
thermal units of heat above that contained by 
1 pound of water at 32° Fahrenheit. This steam 
is condensed to water having a temperature of 
100° to 120°, by mixing it with water of the ordi- 
nary temperature. If the temperature of the in- 
jection water is 50° and that of the hot well 110°, 
then every pound of injection water will take up 
110 — 50=60 units of heat. If the total heat in 
each pound of steam as it leaves the cylinder be 
taken at 1140 above that contained by water at 
32°, then each pound of steam in condensing must 
give up 1140+32—110=1062 units. 
Hence the weight of water to condense j. pound 

„ , 1062 ,„„ 

oi steam= -;rr-= 17.7 pounds. 
bU 

Let H = number of units of heat in 1 pound weight of 

steam above that contained by 1 pound of water 

at 32 degrees. 

T = temperature of hot well. 

t — temperature of injection water. 

W = weight of injection water (in pounds) required 

for each pound of steam. 

H4-32-T 



Then W-- 



T-^ 



134 



MECHANICAL DRAWING 



A common rule is, weight of injection wateri= 
25 to 30 times the weight of the steam to be con- 
densed. 

The volume of a jet condenser should be pro- 
portioned according to the volume of ths low- 
pressure cylinder. The ratio of the volume of the 
condenser to that of the low-pressure cylinder 
varies greatly in practice. It should not be less 
than l^, but we know of examples from recent 
practice where it is as high as l^A. A common rule 
is to make the volume of the condenser half that 
of the low-pressure cylinder exhausting into it. 

The area of the injection orifice is given by the 

iormula A= — . Where 

A=area of valve in square inches, 

W=weight of injection water required per min- 
ute in pounds. 

'y:=velocity of water through the orifice in feet 
per second. 

In many cases in practice v may be taken at 20 

feet per second, then, A= r^ 

Surface Condensers. Since the feed water for 
the boiler is taken from the hot well of the engine, 
it is evident that if the injection water is of a 
character which would be injurious to the boiler 
if pumped into it, there is then a serious objection 
to a jet condenser. The difficulty is got over by 
using a surface condenser. In marine engines 



where the injection water is taken from the sea; 
surface condensers are always adopted. Surface 
condensers are either cylindrical or rectangular 
shells, and are made of brass, cast-iron, wrought- 
iron, or steel. The end or tube plates are gen- 




Fig. 210— Triple Expansion Marine Engine, Side Elevation. 



STEAM ENGINES 



135 



erally made of rolled brass. This form of con- 
denser contains a large number of tubes, which 
pass from one end of the condenser to the other. 
The tubes are made of brass, and are solid drawn, 
and they are sometimes tinned outside and inside. 
They vary in diameter from ^2 to 1 inch, but gen- 
erally they are % inch in diameter outside. The 
thickness of the tubes is from 16 B. W. Gauge to 
19 B. W, Gauge. For %-inch tubes, the pitch, or 
distance from center to center, varies from 1 1/16 
inch to 1% inches. The length of the tubes varies 
greatly, depending on the size and design of the 
condenser. In some cases the length is over 18 
feet. 

Triple Expansion Marine Engines. Figs. 210 
and 211 show elevations of the engines complete, 
which have been prepared from the working 
drawings of the makers. 

The cylinders are placed in the order, high, 
intermediate, low, and their diameters are 8 
inches, 13 inches, and 21 inches respectively. All 
the pistons have a stroke of 16 inches. The cylin- 
ders are in one casting, and are supported at the 
back of the engine on three hollow cast-iron col- 
umns cast on the condenser. At the front the 
cylinders are supported by two turned wrought- 
iron columns, which are flanged at their upper 
ends, and firmly secured at their lower ends to 
the bed of the engine. 



The guides for the crossheads are formed on the 
columns on the condenser. 

The condenser is of cast-iron in one casting, and 
contains 228 tubes, giving 220 square feet of cool- 
ing surface. The tube plates are made of brass. 




Fig. 211— Triple Espansiou Marine Engine, End Elevation. 



136 



MECHANICAL DRAWING 



and are 4 feet 11 inclies apart. The circulating or 
cooling water enters the condenser at the bottom, 
and passes through the lower nest of 114 tubes, 
and back through the upper nest of 114 tubes. 
The cover on one end of the condenser has cast on 
it, on the inside, an air vessel, which serves to 
reduce the shock due to any irregularity in the 
flow of the water which enters at that end. The 
chamber at the same end of the condenser is di- 
vided horizontally by a partition, which is cast 
on the cover and round the air vessel just men- 
tioned. The steam from the low-pressure cylin- 
der passes through the hollow column under it to 
the condenser, where it passes round the tubes 
and is condensed and drawn off at the bottom by 
the air-pump. 



The air-pump is a vertical single-acting bucket 
pump 9 inches in diameter, and the circulating 
pump is a double-acting piston pump 5 inches 
diameter, also vertical. 

There is one bilge pump and on© feed pump, 
each 1% inches in diameter. 

All the pumps have a stroke of 8 inches. 

The crank shaft is forged and in one piece. The 
three cranks are equally^ inclined to one another. 
The main and crank pin bearings are each 4 inches 
in diameter. The thrust is taken by a single collar 
8 inches diameter forced on the shaft. 

The eccentric sheaves are. forged on the crank 
shaft, they are each 7% inches in diameter, and 
have an eccentricity of 1% inches. 



MECHANISMS 



In an important class of mechanisms, it is not 
so much the modification of velocity, as the paths 
which the different points of the mechanism trace 
out, which is of interest. Very frequently some 
point or tool is required to trace out a particular 
curve, such as a circlei, a straight line, an ellipse, 
or a helix; and this has to be done mechanically 
by a combination of links. If the force to be 
transmitted is small, the combination is termed an 
instrument; if considerable, a machine. 

Fi"om a. mechanical point of view, the circle is 
the easiest curve to generate. If a point is re- 
quired tO' move in a circular arc, it need only be 
attached by a link of constant length to a spindle 
which rotates in fixed bearings, as in the ordinary 
crank shaft, arm, and pin arrangement. In the 
case of a circle, therefore, the curve is generated, 
that is to say, a previously existing circle is not 
merely used as a copy. But if any other curve, 
such as a straight line or ellipse, has to be traced 
out, we must either copy a. previously existing 
straight line or ellipse, or else, by means of a 
kinematic chain consisting entirely of turning 
pairs, generate the required curve. We now pro- 



137 



ceed to discuss those mechanisms which either 
generate or copy some particular curve. 

Parallel Motions. By far the most important 
mechanisms are those which either generate or 
copy a straight line. They are usually referred 
to as parallel motions. 

Parallel motions may be divided into two 
classes, namely — 

(1) Those consisting entirely of turning pairs, 
and in which, therefore, a straight line is either 
exactly or approximately generated; 

(2) Those containing one or more sliding 
pairs on which the accuracy of the line traced 
out depends, and which, therefore, may be looked 
upon as copying machines. 

Peaucellier's Cell. Consider, first, those paral- 
lel motions which generate a mathematical 
straight line. The most familiar example is the 
mechanism known as Peaucellier's Cell. It con- 
sists of a jointed rhombus or cell, ABCD, the 
links comprising it being all of the same length 
(Fig. 212). The pins B and C are coupled to a 
fixed center, P^, by two links, PjB and P^C, of 
equal length; while the pin A is coupled to a 



138 



MECHANICAL DEAWING 



second fixed center, Pg, by a link, PgA, equal in 
length to the distance P1P2 between the two cen- 
ters, so that when motion takes place, A is con- 
strained to describe a circular arc passing through 
Pj. Under these conditions, the fourth pin, D, 
will trace out a straight line perpendicular to 
PjPg. It will be noticed that all the links in the 
chain move parallel to one plane. 
To prove this statement, it is clear that in all 




positions of the mechanism the points Pj, A, and 
D are coUinear. Moreover, since the diagonals 
AD, BC of the cell bisect each other at right an- 
gles, 

PiAxPiD=(PiN"— NA) (PiN-f-NA) 

=PiN2— NA2 

=(PiB2— NB2)-^(AB2— NB2) 

=PiB2— AB2; 

so that, sipce the links PjB and AB are of con- 



stant length, it follows that in all positions of the 
mechanism the product PiAxPiD is constant. 

Now let the circle described by A meet the line 
of centers in E, and draw DF perpendicular to the 
line of centers. Since the angles DAE and DFE 
are each a right angle, the quadrilateral DAEF 
may be inscribed in a circle, and consequently the 
product PjExPiF is equal to the product PjAx 
PjD, and is therefore constant. But PiEc=2x 




P2A, and is constant ; hence in all positions of the 
mechanism the distance P^F is constant, from 
which it follows that the locus of D is the line DF. 
In Fig. 212, the center Pj lies outside the rhom- 
bus; but it might equally well lie inside it, as 
shown in Fig. 213. Precisely the same argument 
is applicable to either figure. 

Hart's Parallel Motion. Peaucellier's cell is 
not the only combination of links which generates 



MECHANISMS 



13d 



a true straight line. Any set of links whatever 
which makes the product, PjAxPil^ a constant, 
where Pj is a fixed center, will enable D to trace 
out a straight line, provided A traces out a circle 
which passes through P^. A second combination 
of links satisfying this condition is due to Hart, 
and consists of a crossed parallelogram. In Fig. 
214 the links LM, KN are equal, as also are EX. and 
MN. If the links were opened out, the four links 
would form a Darallelogram ; hence the term 




Fig. 214. 

''crossed parallelogram." It is clear that in all 
positions of the linkage the lines LN and KM are 
parallel; and, moreover, the product LNxKM is 
constant. The latter statement becomes clear if 
NS be drawn parallel to KL, and NT perpendicu- 
lar to KM; hence, since NS=LK=NM, T is the 
middle point of SM. Also LNxKM=KSxKM= 
(KT— TS) (KT+ST)=KT2— ST2=(KN2-NT2) 
— (SN2— NT2) = KN2— SN"2 =. KN^— LK^ = con- 
stant, since KN, LK are links of invariable length. 
Suppose, then, that three points, Pj, A, D (Fig. 



215), lying in one line parallel to either LN or 
KM are taken, so that, in all positions of the me- 
chanism, the points, Pj, A, D will be collinear. 
Then, clearly, PjA is a constant proportion of LN 
and PjD of KM; and consequently, since LNX 
KM is constant, PjAxPiD is constant also. If, 
therefore, P^ be fixed, and, by means of a link, 
PgA, turning about a second fixed center Pg, the 
point A is constrained to describe a circle which 

L N 




Fig. 215. 



passes through Pj, the point D will trace out a 
true straight line, perpendicular to the fixed link 

It will be noticed that Peaucellier's motion con- 
tains eight elements, whilst Hart's contains only 
six. True straight lines may also be generated by 
using higher pairs. 

Scott-Russell Parallel Motion. If, instead of 
generating a true straight line, the object is to 
correctly copy a straight line, or to generate an 



140 



MECHANICAL DRAWING 



approximate straight line, meclianisms consisting 
only of four elements may be used. Thus, for 
example, in the double-slider crank chain shown 
in Fig. 216, there are two blocks connected by a 
link of invariable length — the blocks sliding in 
two perpendicular slots — so that the mechanism 



M^^ 




V///////// 

i 



J^±^ 




Fig. 216. 

consists of two turning and two sliding pairs. If 
C be the central point of the link AB, and P the 
point of intersection of the central lines of the 
slots, it is clear that the length of the line PC is 
equal to either AC or CB, and is therefore con- 
stant. Consequently, as the blocks slide in their 



respective slots, the point C will trace out a circle 
of center P, and the mechanism will not be con- 
strained in any way by adding a crank, PC, turn- 
ing about the center P. In that case one of the 
blocks may be dispensed with, and the mechanism 
shown in Fig. 217 obtained. So long as PC, CA, 
CB are all equal, the point A will describe a 
vertical straight line when B moves along the 




^BJ 1 



Fig. 217. 



horizontal slot. The accuracy of the line traced 
out by A will depend entirely upon the accuracy 
of the line traced out by B, so that this mechan- 
ism is distinctly a copying machine. The block B 
is guided in its straight-line path by contact with 
the surfaces of the slot; but the description of 
how a plane surface is produced is more a matter 
of workshop practice than of kinematics, and it 



MECHANISMS 



141 



will be found in most text-books wMcb. deal witli 
workshop appliances. The mechanism represented 
in Fig. 216 is usually known as the Scott-Russell 
Parallel Motion. 

The Tchebicheff's Parallel Motion. A second 
parallel motion, consisting of four turning pairs, 
which generates an approximate straight line, is 




^7777777777777777777^^ 

Fig. 218. 

Tchebicheff 's. The two levers, P^B and PgD, are 
of equal leng-th, and are crossed, and the coupler 
DB, in the mean position of the mechanism, is 
parallel to the fixed link (Fig, 218). The tracing 
point, A, is the middle point of the coupler, and, 
very approximately, traces out a straight line 
parallel to the fixed link. Thus, in the position 



of the mechanism represented by the dotted lines, 
the central point of the coupler is A^, which prac- 
tically lies on the line BD. The proportions of the 
mechanism will, as in the grasshopper motion, 
depend on the limits of the motion. For example, 
if the coupler is vertical when eitlier lever is 
vertical, and if, then, the tracing point lies exactly 
in the line BD produced (as shown in Fig. 218), 
it may be readily shown that if the fixed link Pi 
Pa be denoted by unity, the length of either lever 
,1 




Fig. 219. 

will be represented by 1.25, and of the coupler by 
0.5. These proportions at once follow froni the 
geometry of the mechanism, and may be left as an 
exercise to the reader. With these proportions, it 
will be noticed that the tracing point is moving in 
a direction exactly parallel to the fixed link in the 
extreme positions, as well as in the mean position, 
of the mechanism. 

Roberts' Parallel Motion, In Roberts' motion 
(Fig. 219) the levers, PjB and PgD, are equal, and 



142 



MECHANICAL DRAWING 



in the mean position of the mechanism the coup- 
ler, BD, is parallel to the fixed link; but, unlike 
Tchebicheff's, the levers are not crossed. The 
tracing point A does not lie in the coupler, but at 
the apex of the isosceles triangle BAD, the apex of 
which lies on the line of centers PiP2- For suffi- 
ciently small displacements, the point A approxi- 
mately traces out the straight line PiP2. More- 
over, if in the extreme positions of the mechanism 
the point A coincides with Pj and Pg, it follows! 




rig. 220. 



that the links BA and AD must be equal in length 
to either of the levers, and that the coupler must 
be half the length of the fixed link. Also, to allow 
this extent of motion, the levers must not be less 
than a certain length, which will depend upon the 
length of the coupler. As a limiting case, when 
A' coincides with Pj, the points B', D' Pg must 
lie on one straight line, and the point D^ will lie 
immediately above the mid-point of PiPg. Under 
these conditions, and remembering that the coup- 



ler is half the length of the fixed link, it may be at 
once shown that the length of the levers must be 
1.186 times the length of the coupler, or 0.593 
times the length of the fixed link. Thus, calling 
the fixed link 1, the length of the coupler is 0.5, 
and the minimum length of the remaining four 
links 0.593. The longer, however, the length of 
the levers, the more accurate the motion. 

The Pantograph. The preceding mechanisms, 
consisting entirely of turning pairs, show how a 




c-^..,.^. 



Fig. 221. 



straight line may be exactly or very approximate- 
ly generated. Scott-Russell's motion is an illus- 
tration of a copying machine, thc' accuracy of the 
line traced out depending upon the accuracy of 
the line copied, and it only truly describes a 
straight line provided the initial line is perfectly 
straight. A' true copying machine is one which 
exactly reproduces the motion of the tracing point 
on a different or on the same scale, all the irregu- 
larities in the original being reproduced in the 



MECHANISMS 



143 



copy. As copying meclianisms are very frequent- 
ly associated with parallel motions, it will be ad- 
visable, at this stage, to consider them. 

The most familiar copying mechanism is the 
pantograph, which may exist in various forms, 
such, for example, as are shown in Figs, 220 to 
223. In each of the first three figures there are 
four links, which form a parallelogram; and if 
the proportions are such that the points 0, A, and 




Fig. 222. 

C are coUinear in one position, they are collinear 
in all positions of the mechanism. In each figure, 
is a fixed center, and the point C traces out a 
curve exactly similar to that traced out by A, 
whatever the shape of that curve may be. Thisi 
follows from the fact that the ratio of the radii 
from the fixed point O to the loci of C and A is 
always constant and equal to 

-^, that is to say, to ^ . In Fig. 221, if D and 



E are the middle points of BC and BA respective- 
ly, the curve traced out by C will be exactly the 
same size as that traced out by A. In Fig. 223 
the links C'G, CE are parallel to AH, AF re- 
spectively. 

The pantograph mechanism is employed to de- 
termine the section of steel rails, tyres, and simi- 
lar bodies, and in engraving machines, beam en- 
gines, indicators, etc. 




Fig. 223. 



Rotary Pumps and Blowers. The method of 
action of a rotary pump will be evident from 
Figs. 224, 225 and 226, which represent a section 
of the pump perpendicular to the axes of rota- 
tion. The curved plates L and M, exactly similar 
in shape, are keyed to the two shafts P^ and Pg, 
which rotate with equal but opposite angular 
velocities in bearings in the casing C. The equiva- 



144 



MECHANICAL DRAWING 



lent pitcii circles of the plates or pistons are rep- 
resented by the two chain-dotted circles of equal 
diameter; and the sides of the casing are at"cs of 
circles having P^ and Pg as centers. The pistons 
are, therefore, always in contact with the casing 
at one or other of the extremities of their major 
axes, and their shape is such that they are always 
in contact with each other. The suction pipe is 
coupled to the casting at the flange S, and the 





Fig. 225. 

delivery pipe at the flange D. As shown in Fig. 
224, the piston- L is rotating in a counter-clock- 
wise direction, and the piston M in a clockwise 
direction. When the shafts have turned through 
a certain angle, the pistons will be as shown in 
Fig. 225. It will be noticed that the space A, 
which in Fig. 226 is in communication with the 
suction pipe, is now entirely cut off, and any fluid 
in it will, neglecting leakage, be carried xound 



with the piston L. After a short time, the upper 
extremity of the major axis ceases to have contact 
with the casing, and the space A is put in com- 
munication with the delivery pipe. No further 
transference of fluid from the suction pipe takes 
place until the position of the pistons is as shown 
in Fig. 226, when the space B is put into com- 
munication with the delivery pipe. Since the 
space above the pistons in Fig. 225 is exactly the 





Fig. 227. 

same as in Fig. 226, it follows that, whatever the 
fluctuations in volume have been in the interval, 
the whole of the fluid which occupied the space 
A has been discharged up the delivery pipe. It 
will thus be seen that the total volume of fluid 
pumped up per revolution is equal tO' four times 
the volume of the space A; or, expressed more 
generally, is equal tO' twice the difference between 
the area of a circle having a diameter equal to the 



MECHANISMS 



145 



major axis of the piston and the area of the pis- 
ton itself, multiplied by the axial length of the 
piston. 

In Fig. 224 each piston has two projections and 
two hollows, but the number of teeth is arbitrary. 
Fig. 227 shows a pump in which the pistons have 
six teeth. As in the previous case, the theoretical 
discharge is equal to the area of the two adden- 
dum circles less the area of the pistons, multi- 
plied by the axial length of the pistons. 

It will be noticed that leakage is prevented by 
the contact of the pistons with the casing and 
with each other. Sliding, and therefore wear, 
takes place not only between the pistons and the 
casing, but also between the pistons themselves, 
and it is a matter of difficulty to keep a fluid-tight 
contact. It is for this reason that, notwithstand- 
ing their great simplicity and the absence of 
valves, they are not often used as water-pumps. 
They are, however, frequently used for supplying 
the blast to furnaces (as in Eoots blowers), and 
also for pumping thick pasty fluids which might 
interfere with the proper working of the valves. 

There are two points of detail which ought to 
be noticed. In the first place, if toothed pistons 
are used, as shown in Fig. 227, one wheel can 
always act as driver to the second; but with the 
pistons shown in Fig. 224 this would be impossi- 
ble in certain portions of the revolution. The 
shafts Pj and Pg almost invariably, therefore, re- 



ceive the equal and opposite rotations from two 
equal spur wheels which gear with each other, 
and are keyed to the shaft on the outside of the 
casing — the wheels being driven, through gear- 
ing, from the engine. The object of the piston is, 
in fact, merely to preserve contact, and not to 
transmit force from one to' the other. In the sec- 
ond place, the circular sides of the casing must be 
rather more than a semi-circumference in length, 
or otherwise communication would be opened with 
the delivery pipe before that with the suction pipe 
was closed. 

Shapes of Blowers. The shapes given to the 
rotatory pistons is arbitrary provided they satis- 
fy the condition that the motion transmitted by 
their sliding contact is exactly the same as that 
which would be transmitted by the rolling contact 
of the two pitch circles; in other words, provided 
the common normal at the point of contact of the 
pistons always passes through the pitch point. 
The point of contact will not always lie on the 
line of centers, and the velocity of sliding at any 
instant will be equal to the length of the common 
normal to the two surfaces, measured to the pitch 
point, multiplied by twice the angular velocity of 
either wheel. 

It has been shown that cycloidal curves satisfy 
the required condition — the parts of the piston 
above the pitch circles being epicycloids, and the 
parts below hypocycloids. The size of the rolling 



146 



MECHANICAL DRAWING 



circle — which, since the pistons are invariably of 
the same shape, will be the same for both the pro- 
jection and the hollow — will depend on the num- 
ber of teeth in the piston, but the diameter of the 
pitch circle must be an exact multiple of the diam- 
eter of the rolling circle. If the pistons are two- 
toothed wheels, as shown in Fig. 224, the diameter 
of the rolling circle will be one-quarter that of 
the pitch circle, and the points where the piston 
profile crosses the pitch circle will divide the lat- 
ter into four equal parts (as in Root's Blower). 




If the pistons are six-toothed wheels, as shown in 
Fig. 227, the diameter of the rolling circle is one- 
twelfth that of the pitch circle; and so on. Fig. 
228 shows a pair of pistons consisting of two half- 
cylinders of different diameters and connected by 
the curves 1 and m. The circular portions roll 
without slipping, and the curves 1 and m must 
satisfy the necessary kinematic condition. If the 
portion below the pitch circles be a radial straight 
line, the portion above it must be an epicycloid 



formed by a rolling circle half the size of the pitch 
circle. 
The profiles need not, however, consist of 




curves possessing any particular mathematical 
properties. If the profile of one of the pistons is 



MECHANISMS 



147 



given, that of the second can be found by the 
geometrical or mechanical method. 

For example, Fig. 229 shows an illustration of 
the geometrical method in which the points of 
each piston are circular 'arcs subtending an angle 
of 90° at the center — the height of the point being 
taken as half the radius of the pitch circle. The 
center of the circle forming the point of the piston 
keyed to Pg is C, and, following the method a 
number of normals through the points a, b, c, . . . 
are drawn to the circular arc. These all neces- 
sarily pass through the point C, and they inter- 
sect the pitch circle B of center Pg in the points 

1, 2, 3,. . . With K (the pitch point) as center, 
arcs passing through the points 1, 2, 3, . . . are 
struck to meet the pitch circle A in the points 1, 

2, 3, . . . This construction, since the pitch circles 
are equal, merely makes the arcs Kl, 1 2, 2 3 . . . 
on each pitch circle the same. With the points on 
A as centers, and radii equal to la, 2b, 3c, . . . cir- 
cular arcs are drawn; and the envelope of these 
arcs is the proper shape of the hollow of the 



piston on P to gear with the given point. The 
whole pistons are as shown. Other shaped pistons, 
obtained in the same way, are shown in Fig. 230. 

In the mechanical method, a template of one 
piston would be cut out, and the two pitch circles 
would roll together as explained. The profile of 




rig. 230. 

the given piston would be marked off on the sheet 
rotating about the second center; and the envel- 
ope of the curves so obtained would be the shape 
of the second piston. In Fig. 230 the shape of the 
second piston is governed by the path of the ex- 
treme points of the first piston. 



WORK DONE BY A DROP HAMMER 



The great value of the drop hammer is its sim- 
Dlicity and cheapness as a means of storing energy, 
which can be given out again in doing the work of 
raising, stamping, and forging. 

Ttie drop hammer is an example of the useful 
application of the principle of the falling weight. 
In finding the work accumulated in any moving 
body^ such, for instance, as energy stored up in 
a flywheel, the work of a locomotive when ascend- 
ing an incline, the work of a cannon ball, and 
similar questions, it is necessary to introduce the 
force of gravity into the calculation, since the 
law of gravitation necessarily has an effect upon 
such bodies as are dealt with in these calculations. 
It is therefore perhaps necessary to briefly review 
the action of gravity on falling bodies to enable 
the practical mechanic to understand what the 
gravity constant means. 

If a body be raised 16 1/12 feet, then allowed 
to fall- freely, it will fall through this space of 
16 1/12 feet in one second, and at the end of that 
second it will have attained a velocity of 32 1/6 
feet per second. This velocity of 32 1/6 feet per 
second, wMch is simply due to the force of gravity, 
is denoted by g and the velocity v attained at 
the end of t seconds will be tx32 1/6, or v=tx 
32 1/6. Therefore the velocity of a body which 
has fallen for four seconds will be at the end of 



that time travelling at a velocity=4X32 1/6=128 
2/3 feet per second. 

The mean velocity — that is, the velocity in the 
middle of that time— will be 2x32 1/6=64 1/3 
feet per second, and the space described will be 
4^x16 1/12=257 1/3 feet. 

Formula for Falling Bodies when Falling from 

Rest. 

h = height of fall in feet, 
V ^ velocity in feet per second, 
g = force of gravity = 32.2. 
t = time of fall in seconds. 

2h 



v=gt = — =i/2gh 
~g V \ g 

If the drop hammer be raised to a definite 
height, the work expended in raising it will be 
Wxh, and in doing so the force of gravity will 
have to be overcome. If the hammer be now sup- 
ported, there will be potential energy stored up in 
it, and when allowed to fall it will attain a certain 
velocity depending upon the distance fallen. 
When the hammer reaches the end of its fall, the 
accumulated work in the hammer will be 



lis 



wv 

2g 



= Wh 



WORK DONE BY A DROP HAMMER 



149 



Example. Suppose a drop hammer of 500 
pound weight be raised through a height of 579 
feet. The work expended in raising this hammer 
will be 

Wxh=500x 579=289500 foot-pounds. 
If the hammer be now supported at this height, 
the potential energy which exists iji, or is stored 
up, will be::=289,500 foot-pounds. When allowed 
to fall the accumulated work will be equal to 

Wv2 



First find v: 



2g 



v=i/2gh=v'2X32iX579=193 feet per second, 
when the hammer reaches the end of its fall it 
will have attained a velocity equal to 193 feet per 
second, therefore 

Wv2 500X193X193 „„^^^„ , , 
"2^= — 2X32' =289500 foot-pounds, 

and this is the same result as Wxh. 

A certain amount of energy is passed into the 
drop hammer in raising it up, and when it falls 
the energy is given out again. There is neither 
gain nor loss of power. It may be that a great 
pressure is exerted through a small space, or a 
less pressure through a greater space, and in both 
instances the work may be the same. 

If a resistance is offered to a 560-pound ham- 
mer, falling from a height of 16 feet, during the 
last one foot of its fall the average pressure acting 
against the resistance will be 8,960 pounds, the 



pressure being much greater at the commence- 
ment, and reducing as it reaches the last inch, the 
accumulated energy gradually decreasing to sim- 
ply that of the weight of the hammer itself as it 
reaches the end of its fall. 

But should the hammer be brought to rest in a 
fraction of 1 foot, then the resistance offered must 
be proportionately greater. 

A stamp hammer 200 pounds in weight falls 10 
feet, and in stamping a piece of metal the hammer 
is brought to rest in the space of the last % inch 
of its fall. What resistance has been offered by 
the metal article? 

WXh=200XlO=2000 foot-pounds, 
i inch = i of tV = T4 of 1 foot, 
RXl 



and since 2000= 



25 



2000X24 
therefore R= z =48000 pounds. 

Work done by a hand hammer. The conditions 
under which the hand hammer is used make it 
necessary that the law of gravitation shall be 
introduced into the calculation. Take the case 
of a machinist striking a blow upon the head 
of a chisel, or driving a nail into a piece of wood, 
with a 2-pound hand hammer. 

As another example, consider the case of a ma- 
chinist driving a key into a boss of a flywheel 
with a 4-pound hammer. In the first case there 
are two forces acting upon the hammer, namely, 



150 



MECHANICAL DRAWING 



force of gravity and tlie man's muscular force. 
The workman raises the hammer, he then drives 
it home, delivering a blow upon the head of the 
chisel. The first portion of the distance through 
which the hammer moves is traversed by a move- 
ment of the whole arm from the shoulder. 

This is followed up by the workman straighten- 
ing his arm at the elbow, then, just as he is 
about to reach the head of the chisel with the ham- 
mer to strike the blow, he straightens his wrist, 
thereby adding impetus to the hammer, which is 
already rapidly falling, and in this manner a very 
great velocity is given, probably at the exact mo- 
ment of impact the actual velocity may be 50 feet 
per second. 

In the second example, where a blow is deliv- 
ered upon the head of a steel key, by a hammer, 
and the hammer is driven in a horizontal line, the 
machinist will swing the hammer through a com- 
paratively long distance, and will probably put 
the weight of the upper half of his body into the 
blow, thereby considerably increasing the velocity 
of the hammer, which may be 50 feet per second, 
as before. In both these cases of hand hammers 
the accumulated work or energy stored up in the 
hammer will be the same as though the hammer 
had fallen from a sufficient height to attain that 
velocity which the hammer has at the moment 
of impact. But here the only information we have 
to assist us in solving the problem is the weight 
of the hammer and the assumed velocity at which 



it is moving, say 50 feet per second. Since hav- 
ing no particulars as to the height from which a 
body must fall to attain this velocity, it is neces- 
sary to introduce the law of gravitation into the 
calculation to enable reliable results to be ob- 
tained. 

We have for the 2-pound hammer accumulated 
work 

Wv2_2X50X50 „ 

"22" 2X32 ~ foot-pounds. 

If the face of the hammer moves the head of the 
chisel 1/16 inch, then 

T^X^-ila of 1 foot, 

78X192 
and R= r =14976 pounds. 

If a nail had been driven i/4 inch, 

tXtV=tV of 1 foot, 

78X48 
and K= — - — =3744 pounds. 

In the case of the 4-pound hammer w© have 

accumulated work 

Wv2 4X50X50 _„ , 

~2s~ 2X32 ^-^^^ foot-pounds. 

If the key is driven % inch by the blow, 

iXTV=inr of 1 foot. 

156X96 
and R= — z =14976 pounds resistance. 

The work done by the jack hammer, namely, 
14,976 pounds, is approximately the same that 
would be obtained by a dead load of 14,976 pounds 
giving a direct pressure. 



CIRCUMFERENCES AND AREAS OF CIRCLES. 



151 







Circumference of 


Circles. 






■ 




Area of Circles. 






Diam. 


Circmn. 


Diam. 


Circmn. 


Diam. 


Circum. 


Diam. 


Circum. 


Diam. 


Area. 


Diam. 


Area. 


Diam. 


Area. 


Diam. 


Area. 




.3927 


10 


31.41 


30 


94,24 


65 


204.2 


H 


0.0123 


10 


78.54 


30 


706.86 


65 


3318.3 


.7854 


lOX 


32.98 


31 


97.38 


66 


207.3 




14 


0.0491 


101/2 


86.59 


31 


754.76 


66 


3421.2 


y% 


1.178 


11 


34.55 


32 


100.5 


67 


210.4 




% 


0.1104 


11 


95.03 


32 


804.24 


67 


3525.6 


H 


1.570 


UK 


3ti.l2 


33 


103.6 


68 


213.6 




Vi 


0.1963 


11^ 


103.86 


33 


855.30 


68 


3631.6 


% 


1.963 


12 


37.69 


34 


106.8 


69 


216.7 




% 


0.3068 


12 


113.09 


34 


907.92 


69 


3739.2 


H 


2.356 


12>^ 


39.27 


35 


109.9 


70 


219.9 




3/4 


0.4418 


12^2 


122.71 


35 


962.11 


70 


3848.4 




2.748 


13 


40.84 


36 


113.0 


71 


223.0 




^ 


0.6013 


13 


132.73 


36 


1017.8 


71 


3959.2 


3.141 


13>^ 


42.41 


37 


116.2 


72 


226.1 




1 


0.7854 


13^/2 


143.13 


37 


1075.2 


72 


4071.5 


'i-Vs 


3.534 


14 


43.98 


38 


119.3 


73 


229.3 




11/8 


0.9940 


14 


153.93 


38 


1134.1 


73 


4185.4 


1^ 


3.927 


14>^ 


45.55 


39 


122.5 


74 


232.4 




11/4 


1.227 


141/2 


165.13 


39 


1194.5 


74 


4300.8 


w& 


4.319 


15 


47.12 


40 


125.6 


75 


235.6 




1% 


1.484 


15 


176.71 


40 


1256.6 


75 


4417.8 


iH 


4.712 


15^ 


48.69 


41 


128.8 


76 


238.7 




11/2 


1.767 


15^2 


188.69 


41 


1320.2 


76 


4536.4 


5.105 


16 


50.26 


42 


131.9 


77 


241.9 




15/8 


2.073 


16 


201.06 


43 


1385.4 


77 


4656.6 


134 


5.497 


16K 


51.83 


43 


135.0 


78 


245.0 




134 


2.405 


16^2 


213.82 


43 


1452.2 


78 


4778.3 


IH 


5.890 


17 


53.40 


44 


138.2 


79 


248.1 




17/8 


2.761 


17 


226.98 


44 


1520.5 


79 


4901.6 


2 


6.283 


17>^ 


54.97 


45 


141.3 


80 


251.3 




2 


3.141 


17^2 


240.52 


45 


1590.4 


80 


5026.5 


2X 


7.068 


18 


56.54 


46 


144.5 


81 


254.4 




214 


3.976 


18 


254.46 


46 


1661.9 


81 


5153.0 


^'A 


7.854 


18>^ 


58.11 


47 


147.6 


82 


257.6 




21/2 


4.908 


181/2 


268.80 


47 


1734.9 


82 


5381.0 


234 
3 


8.639 


19 


59.69 


48 


150.7 


83 


260.7 




234 


5.939 


19 


283.52 


48 


1809.5 


83 


5410.6 


9.424 


19J^ 


61.26 


49 


153.9 


84 


263.8 




8 


7.068 


191/2 


298.64 


49 


1885.7 


84 


5541.7 


B'X 


10.21 


20 


62.83 


50 


157.0 


85 


267.0 




314 


8.295 


20 


314.16 


50 


1963.5 


85 


5674.5 


S'A 


10.99 


20 J^ 


64.40 


51 


160.2 


86 


270.1 




ZV2 


9.621 


201/2 


330.06 


51 


2042.8 


86 


5808.8 


334 

4 


11.78 


21 


65.97 


52 


163.3 


87 


273.3 




334 


11.044 


21 


346.36 


52 


2123.7 


87 


5944.6 


12.56 


21 K 


67.54 


53 


166.5 


88 


276.4 




4 


12.566 


2II/2 


363 05 


53 


2206.1 


88 


6082.1 


4K 


14.13 


22 


69.11 


54 


169.6 


89 


279.6 




4H 


15.904 


22 


380.13 


54 


2290.2 


89 


6331.1 


5 


15.70 


221/2 


70.68 


55 


172.7 


90 


282.7 




5 


19.635 


221/2 


397.60 


55 


2375.8 


90 


6361.7 


5'A 


17.27 


23 


72.25 


56 


175.9 


91 


285.8 




51/2 


23.758 


23 


415.47 


56 


2463.0 


91 


6503.9 


6 


18.84 


231/2 


73.82 


57 


179.0 


92 


289.0 




6 


28.274 


23H 


433.73 


57 


2551.7 


92 


6647.6 


6K 


20.42 


24 


75.39 


58 


181.2 


93 


292.1 




6I/2 


33.183 


24 


452.39 


68 


2642.0 


93 


6793.9 


7 


21.99 


241/2 


76.96 


59 


185.3 


94 


295.3 




7 


38.484 


241/2 


471.43 


59 


2733.9 


94 


6939.8 


"^'A 


23.56 


25 


78.54 


60 


188.4 


95 


298.4 




7H 


44.178 


25 


490.87 


60 


2827.4 


95 


7088.3 


8 


25.13 


26 


81.68 


61 


191.6 


96 


301.5 




8 


50.265 


26 


530.93 


61 


2922.4 


96 


7238.2 


8K 


26.70 


27 


84.82 


62 


194.7 


97 


304.7 




8^2 


56.745 


27 


572.55 


62 


3019.0 


97 


7389.8 


9 


28.27 


28 


87.96 


63 


197.9 


98 


307.8 1 


9 


63.617 


28 


615.75 


63 


3117.2 


98 


7542.9 


9>^ 


29.84 


29 


91.10 


64 


201.0 


99 


311.0 




9H 


70.882 


29 


660.52 


64 


3216.9 


99 


7697.7 



152 



MECHANICAL DRAWING 



To compute tlie circmnference of a diameter 
gTeater tliaii any in the above table: 

Eule. — Divide tlie dimension by 2, 3, 4, etc., if 
practicable, mitil it is reduced to a diameter to be 
found in table. Take the tabular circumference 
of this diameter, multiply it by 2, 3, 4, etc., accord- 
ing as it was divided, and the product will be the 
circumference required. 

Example. — What is the circumference of a di- 
ameter of 125? 125-f-5=25. Tabular circumfer- 
ence of 25=78.54, 78.54X5=392.7, circumference 
required. 



To compute the area of a diameter greater than 
any in the above table: 

Eule. — Divide the dimension by 2, 3, 4, etc., if 
practicable, until it is reduced to a quotient to be 
found in the table, then multiply the tabular area 
of the quotient by the square of the factor. The 
product will be the area required. 

Example. — What is area of diameter of 150? 
150^5=30. Tabular area of 30=706.86 which X 
25=17,671.5 area required. 



LOGARITHMS OF NUMBERS 



Logaritlims are tlie exponents of a series of 
powers and roots of numbers. Ttie logarithm of 
a number is that exponent of some other number, 
which renders the power of the other number, 
which is denoted by the exponent, equal to the 
former. In other words the logarithm of a num- 
ber is the exponent of the power to which the 
number must be raised to give a. given base. 

When the logarithms of numbers form a series 
in arithmetical progression, their corresponding 
natural numbers form a series in geometrical pro- 
gression, thus: 

Common logarithms 12 3 4 

Natural numbers 1 10 100 1,000 10,000 

Natural logarithms were the invention of Lord 
Napier. Common logarithms, the kind in general 
use, were invented by Prof. Briggs of Oxford, 
Elngland. Logarithms are extremely useful in 
shortening the labor of mathematical calculations. 

The addition and subtraction of common loga- 
rithms correspond to the multiplication and di- 
vision of their natural numbers. 

In a like manner, involution is performed by 
multiplying the common logarithm of any number 
by the number denoting the required power, and 



153 



evolution by dividing the common logarithm of 
the number denoting the required root. 

The common logarithm of a number consists of 
two parts, an integral part or whole number, 
which is called the characteristic, and a decimal 
called the mantissa. 

To find the common logarithm of a given num- 
ber from the Table proceed as follows: 

The first two figures of the number will be found 
in the vertical column to the extreme left in the 
table, and the third figure of the number in the 
horizontal row at either the top or bottom of 
the table. Having found the first two figures of the 
number, always neglecting the decimal, pass along 
the line opposite these figures until the column 
headed by the third figure of the number is 
reached. The number thus found will be the man- 
tissa or decimal fraction of the logarithm. The 
characteristic will depend upon the number of 
integers or whole numbers, less one, in the num- 
ber, counting from the left of the decimal point. 
If the decimal point be entirely to the left of the 
number, the characteristic is obtained by counting 
the number of cyphers before the first number, to 
the right and adjacent to the decimal point. 



154 



MECHANICAL DRAWING 



Kxample: Find tlie common logarithm of 5.06 
from the Table. 

Answer: In the row of figures opposite 50 and 
in the column under 6, the mantissa of the loga- 
rithm is .7042. Counting from the decimal place 
of the number to the left, the characteristic will be 
one less than the number of figures to the right of 
the decimal point, which is, in this case, 1, and 1 
minus 1 equals zero, which is the characteristic of 
the mantissa .7042, the complete logarithm of 5.06 
will then be 0.7042. 

The logarithm of 0.506 is— 1.7042 
The logarithm of 5.06 is 0.7042 
The logarithm of 50.6 is 1.7042 
The logarithm of 506 is 2.7042 

To find the number corresponding to a given 
logarithm: As the mantissa, of the given loga- 
rithm is not usually found in the table, select the 
four figures corresponding nearest to the given 
mantissa. The first two figures of the number will 
be found in the column marked ' ' No. ' ' at the left 
of the row in which is the mantissa elected, and 
the third or last figure of the number at the top or 
bottom of the vertical row of figures. 

Example: Find the number from the Table 
corresponding to logarithm 1.0334. 

Answer: The first two figures of the number 
corresponding to the mantissa .0334 are 10, and 



at the top of the vertical column the third figures 
given as 8, making the three figures 108. As the 
characteristic is 1 therefore the actual number is 
10.8. 

The number corresponding to — 1.0334 is .108 
The number corresponding to 0.0334 is 1.08 
The number corresponding to 1.0334 is 10.8 
The number corresponding to 2.0334 is 108 

To multiply one or more numbers together, add 
the common logarithms of the numbers together, 
the sum will be the logarithm of the required num- 
ber. 

To divide a number by one or more numbers, 
subtract the suma of the common logarithms of the 
numbers from the logarithms of the number to 
be divided. 

The mantissa of the common logarithm of 6 is 
the same as the mantissa of 60 or 600, the char- 
acteristic only being changed thus: 

Common logarithm of .600=— 1.7782 
Common logarithm of 6.00= 0.7782 
Common logarithm of 60.0= 1.7782 
Common logarithm of 600 = 2.7782 

The table gives the common logarithms of num- 
bers from 100 to 999. 

Note. A decimal point must always be pre- 
fixed to the mantissa of a logarithm obtained from 
the table, before affixing the characteristic. 



LOGARITHMS OF NUMBERS 



155 



Logarithms of Numbers prom 100 to 999. 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Diff. 


10 

11 

12 
13 

14 
15 
16 

17 
18 
19 

20 

21 
22 
23 

24 
25 
26 

27 
28 
29 

30 

31 
32 
33 


0000 

0414 
0792 
1139 

1561 
1761 
2041 

2304 
2553 

2788 

3010 

3222 
3424 
3617 

3802 
3979 
4150 

4314 

4472 
4624 

4771 

4914 
5051 
5185 


0043 

0453 
0828 
1173 

1492 
1790 
2068 

2330 

2577 
2810 

3032 

3243 
3444 
3636 

3820 
3997 
4166 

4330 

4487 
4639 

4786 

4928 
5065 
5198 


0086 

0492 
0865 
1206 

1523 
1818 
2095 

2355 
2601 
2833 

3054 

3263 
3464 
3655 

3838 
4014 
4183 

4346 
4502 
4654 

4800 

4942 
5079 
5211 


0128 

0531 
0899 
1239 

1553 

1847 
2122 

2380 
2625 
2856 

3075 

3284 
3483 
3674 

3856 
4031 
4200 

4362 
4518 
4669 

4814 

4955 
5092 
5224 


0170 

0569 
0934 
1271 

1548 
1875 
2148 

2405 
2648 
2878 

3096 

3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 

4829 

4969 
5105 
5237 


0212 

0607 
0969 
1303 

1614 
1903 
2175 

2430 
2672 
2900 

3118 

3324 
3522 
3711 

3892 
4065 
4232 

4393 

4548 
4698 

4843 

4983 
5119 
5250 


0253 

0645 
1004 
1335 

1644 
1931 
2201 

2455 
2695 
2923 

3139 

3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 

4857 

4997 
5132 
5263 


0294 

0682 
1038 
1367 

1673 
1959 

2227 

2480 
2718 
2945 

3160 

3365 
3560 

3747 

3927 
4099 
4265 

4425 
4579 
4728 

4871 

5011 
5145 
5276 


0334 

0719 
1072 
1399 

1703 
1987 
2253 

2504 
2742 
2967 

3181 

3385 
3579 
3766 

3945 
4116 
4281 

4440 
4594 
4742 

4886 

5024 
5159 
5289 


0374 

0755 
1106 
1430 

1732 
2014 
2279 

2529 
2765 
2989 

3201 

3404 
3598 
3784 

3962 
4133 
4298 

4456 
4609 

4757 

4900 

5038 
5172 
5302 


40 

37 
33 
31 

29 

27 
25 

24 
23 

22 

21 

20 
19 
18 

17 
17 
16 

16 
15 
14 

14 

13 
13 
13 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Diff, 



Logarithms of Numbers from 100 to 999. — (Cont.) 



No. 



34 
35 
36 

37 
38 
39 

40 

41 
42 
43 

44 
45 
46 

47 
48 
49 

50 

51 
52 
53 

54 
55 
56 



No. 



5315 
5441 
5563 

5682 
5798 
5911 

6021 

6128 
6232 
6335 

6435 
6532 
6628 

6721 
6812 
6902 

6990 

7076 
7160 
7243 

7324 
7404 
7482 



5328 
5453 
5575 



5694 
5809 
5922 

6031 

6138 
6242 
6345 

6444 
6542 
6637 

6730 
6821 
6911 

6998 

7084 
7186 
7251 

7332 
7412 
7490 



5340 
5465 
5587 



5705 
5821 
5933 

6042 

6149 
6253 
6355 

6454 
6551 
6646 



5353 
5478 
5599 

5717 
5832 
5944 

6053 

6160 
6263 
6365 

6464 
6561 
6656 



6739 6749 



6830 
6920 

7007 

7093 
7177 
7259 

7340 
7419 
7497 



6839 
6928 

7016 

7101 
7185 
7267 

7348 

7427 
7505 



5366 
5490 
5611 

5729 
5843 
5955 

6064 

6170 
6274 
6375 

6474 
6571 
6665 

6758 
6848 
6937 

7024 

7110 
7193 
7275 

7356 
7435 



5378 
5502 
5623 

5740 
5855 
5966 

6075 

6180 
6284 
6385 

6484 
6580 
6675 

6767 
6857 
6946 

7033 

7118 

7202 
7284 

7364 
7443 



7512 7520 



5391 
5514 
5635 

5752 
5866 
5977 

6085 

6191 
6294 
6395 

6493 
6590 
6684 

6776 
6866 
6955 

7042 

7126 
7210 
7292 

7372 
7451 
7528 



5403 5416 5428 
5527 55395551 
5647 5658:5670 



5763 

5877 
5988 

6096 

6201 
6304 
6405 

6503 
6599 
6693 

6785 
6875 
6964 

7050 

7135 
7218 
7300 

7380 
7459 
7536 



57755786 
58885899 
59996010 

61076117 



6212 
6314 
6415 

6513 
6609 
6702 

6794 
6884 
6972 

7059 

7143 
7226 



7308 7316 



7388 
7466 
7543 



6222 
6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 

7067 

7152 
7235 



7396 
7474 
7551 



Diff. 



13 
12 
12 

12 
12 
12 

11 

10 
10 
10 

10 

10 

9 

9 
9 
9 



Diff. 



156 



MECHANICAL DRAWING 



Logarithms of Numbers from 100 to 999. — (Cont.) 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Diff. 


57 
58 
59 

60 

61 
62 
68 

64 
65 
66 

67 
68 
69 

70 

71 
72 
73 

74 
75 
76 

77 
78 
79 


7559 
7634 
7709 

7782 

7853 
7924 
7993 

8062 
8129 
8195 

8261 
8325 
8388 

8451 

8513 
8573 
8633 

8692 
8751 
8808 

8865 
8921 
8976 


7566 
7642 
7716 

7789 

7860 
7931 
8000 

8069 
8136 
8202 

8267 
8331 
8395 

8457 

8519 
8579 
8639 

8698 
8756 
8814 

8871 
8927 
8982 


7574 
7649 
7723 

7796 

7868 
7938 
8007 

8075 
8142 
8209 

8274 
8338 
8401 

8463 

8525 
8585 
8645 

8704 
8762 
8820 

8876 
8932 
8987 


7582 
7657 
7731 

7803 

7875 
7945 
8014 

8082 
8149 
8215 

8280 
8344 
8407 

8470 

8531 
8591 
8651 

8710 
8768 
8825 

8882 
8938 
8993 


7589 
7664 
7738 

7810 

7882 
7952 
8021 

8089 
8156 

8222 

8287 
8351 
8414 

8476 

8537 
8597 
8657 

8716 
8774 
8831 

8887 
8943 
8998 


7597 
7672 
7745 

7818 

7889 
7959 
8028 

8096 
8162 
8228 

8293 
8357 
8420 

8482 

8543 
8603 
8663 

8722 
8779 
8837 

8893 
8949 
9004 


7604 
7679 

7752 

7825 

7896 
7966 
8035 

8102 
8169 
8235 

8299 
8363 
8426 

8488 

8549 
8609 
8669 

8727 
8785 
8842 

8899 
8954 
9009 


7612 
7686 
7760 

7832 

7903 
7973 
8041 

8109 
8176 
8241 

8306 
8370 
8432 

8494 

8555 
8615 
8675 

8733 
8791 
8848 

8904 
8960 
9015 


7619 
7694 
7767 

7839 

7910 
7980 
8048 

8116 
8182 
8248 

8312 
8376 
8439 

8500 

8561 
8621 
8681 

8739 
8797 
8854 

8910 
8965 
9020 


7627 
7701 

7774 

7846 

7917 
7987 
8055 

8122 
8189 
8254 

8819 
8382 
8445 

8506 

8567 
8627 
8686 

8745 
8802 
8859 

8915 
8971 
9025 


7 
8 
8 

7 

7 
6 
7 

7 
6 
8 

6 
6 
6 

7 

6 
6 
6 

6 
6 
6 

6 
5 
6 


No. 





1 


2 


8 


4 


5 


6 


7 


8 


9 


Diff. 



Logarithms of Numbers from 100 to 999. — (Cont.) 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Diff. 


80 

81 
82 
83 

84 
85 
86 

87 
88 
89 

90 

91 
92 
93 

94 
95 
96 

97 
98 
99 


9031 

9085 
9138 
9191 

9243 
9294 
9345 

9395 

9445 
9494 

9542 

9590 
9638 
9685 

9731 

9777 
9823 

9868 
9912 
9956 


9036 

9090 
9143 
9196 

9248 
9299 
9350 

9400 
9450 
9499 

9547 

9595 
9643 
9689 

9736 

9782 
9827 

9872 
9917 
9961 


9042 

9096 
9149 
9201 

9253 
9304 
9355 

9405 
9455 
9504 

9552 

9600 
9647 
9694 

9741 
9786 
9832 

9877 
9921 
9965 


9047 

9101 
9154 
9206 

9258 
9309 
9360 

9410 
9460 
9509 

9557 

9605 
9652 
9699 

9745 
9790 
9836 

9881 
9926 
9969 


9053 9058 

9106 9112 
9159 9165 
9212 9217 

9263 9269 
9315 9320 
9365 9369 

9415 9420 
9465 9469 
9513 9518 

9562 9566 

9609 9614 
9657 9661 
9703 9708 

9750 9754 
97959800 
9841 9845 

9886^9890 
99309934 
9974|9978 


9063 

9117 
9170 
9222 

9274 
9325 
9374 

9425 
9474 
9523 

9571 

9619 
9666 
9713 

9759 
9805 
9850 

9894 
9939 
9983 


9069 9074 

9122 9128 
9175 9180 
9227 9232 

9279 9284 
9330 9335 
93799384 

9430 9435 
9479,9484 
9528 9533 

9576 9581 

9624 9628 
9671 9675 
9717 9722 

9763 9768 
9809 9814 
9854 9859 

9899 9903 
9943 9948 
9987 9991 


9079 

9133 
9186 
9288 

9289 
9340 
9389 

9440 
9489 
9538 

9586 

9633 
9680 
9724 

9773 
9818 
9863 

9908 
9952 
9996 


6 

5 
5 
5 

5 
5 
5 

5 
5 

4 

4 

5 

5 
4 

4 
5 
5 

4 
4 
4 


No. 





1 


2 


3 


4 j 5 


6 


7 


8 


9 


Diff. 



INDEX 



PAGE 

Preface 5 

Drafting Tools 7 

Geometrical Definitions of Plane Figures .... 20 

Definition of Polygons 25 

Geometrical Definition of Solids 25 

Mechanical Drawing 27 

Geometrical Problems 28 

Mensuration of Plane Surfaces 36 

Mensuration of Volume and Surface of Solids 39 

Mensuration of Triangles 42 

The Mechanical Powers 44 

The Development of Curves 52 

The Development of Surfaces 57 



PAGE 

The Intersection of Surfaces 63 

General Instructions for Machine Drawing ... 67 

Machine Drawing 77 

Technical Definitions 97 

Materials Used in Machine Construction 99 

Machine Design 103 

Horsepower of Gears 115 

Steam Boilers 121 

Steam Engines 127 

Mechanisms 137 

Work Done by a Drop Hammer 148 

Circumferences and Areas of Circles 151 

Logarithms of Numbers 153 



157 



SHEET METALAyORKERSi 
•INSTRU0OR:V: i 




Sheet 
Metal 
Workers'* 
Instructor 



By 
Joseph H. Rose 



Profusely Illustrated. 

T^HIS work consists of useful information for Sheet Metal 
* Workers in all branches of the industry, and contains 
practical rules for describing the various patterns for sheet 
iron, copper and tin work. Geometrical construction of 
plane figures. Examples of practical pattern drawing. 
Tools and appliances used in sheet metal work. Examples 
of practical sheet metal work. Geometrical construction 
and development of solid figures. Soldering and brazing. 
Tinning. Retinning and galvanizing. Materials used in 
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lavatory. A modern bath room. Bath tubs. Lavatories. 
Closets. Urinals. Laundry tubs. Shower baths. Toilet 
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THE AUTOMOBILE HAMOK 



A WORK of practical information for the use of Owners, Operators and 
Automobile Mechanics, giving full and concise information on all 
questions relating to the construction, care and operation of gasoline and 
electric automobiles, including Road Troubles, Motor Troubles, Carbu- 
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iges. With numerous tables, useful 
lies and formulas, wiring diagrams and 
_ /er 100 illustrations, by 

L. ELLIOTT BROOKES 

Aothor o( the "ConstrnctioD of a Gasoline Motor." 

'"■ — t-"™ — =^ 



^»"-^ 
"5^ 



This work has been written especially 
for the practical information of automo- 
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In presenting this work to readers who may be interested in automo* 
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The perusal of this work for a few minutes when troubles occur, will 
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The best and latest and most 
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book is intended for the practical 
instruction of Machinists, Engin- 
eers and others who are interested 
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modern machine shop. The first 
portion of the book is devoted to 
practical esamplesin Arithmetic, 
Decimal Fractions, Roots of Num- 
bers, Algebraic Signs and Symbols, 
Reciprocals and Logarithms of 
Numbers, Practical Geometry and 
and Mensuration. Also Applied 
Mechanics — which includes: The 
lever. The wheel and pinion. The 
pulley. The inclined planes. The 
wedge The, screw and safety valve 
— Specific gravity and the velocity 
of falling bodies — Friction, Belt 
Pulleys and Gear wheels. 

Properties of steam. The Indi- 
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The latter part of the book gives full and complete information 
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Shop tools, Machine tools, Boring machines, Boring mills. Drill 
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